NCERT Solutions for Class 12 Maths Chapter 5: Exercise Questions
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NCERT Continuity and Differentiability Class 12 Solutions: Exercise: 5.1
Page number: 116-118
Total questions: 34
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Question 1: Prove that the function $f ( x) = 5 x -3$ is continuous at $x = 0, at\: \: x = - 3$ and at $x = 5$
Answer:
The given function is
$f ( x) = 5 x -3$
$f(0) = 5(0)-3 = -3$
$\lim\limits_{x\rightarrow 0} f(x) = 5(0)-3 = -3$
$\lim\limits_{x\rightarrow 0} f(x) =f(0)$
Hence, the function is continuous at x = 0
$f(-3)= 5(-3)-3=-15-3=-18 \Rightarrow \lim\limits_{x \rightarrow -3} f(x) = 5(-3)-3 = -15-3=-18$
$ \Rightarrow \lim\limits_{x \rightarrow -3} f(x) = f(-3)$
Hence, the function is continuous at $x = -3$
$f(5)= 5(5)-3=25-3=22 \Rightarrow \lim\limits_{x \rightarrow 5} f(x) = 5(5)-3 = 25-3=-22$
$\Rightarrow \lim\limits_{x \rightarrow 5} f(x) = f(5)$
Hence, the function is continuous at $x = 5$
Question 2: Examine the continuity of the function $f (x) = 2x ^2 - 1 at x = 3.$
Answer:
The given function is
$f(x) = 2x^2-1$
at $x = 3$
$f(3) = 2(3)^2-1 = 2\times 9 - 1=18-1=17\\$
$ \lim\limits_{x\rightarrow 3}f(x) = 2(3)^2-1=2\times 9-1=18-1=17$
$\lim\limits_{x\rightarrow 3}f(x) = f(3)$
Hence, the function is continuous at $x = 3$
Question 3: Examine the following functions for continuity.
$(a) f (x) = x - 5$
Answer:
The given function is
$f(x) = x-5$
Our function is defined for every real number, say k
and value at $x = k$, $f(k) = k-5$
And also,
$\lim\limits_{x\rightarrow k} f(x) = k -5\\$
$ \lim\limits_{x\rightarrow k} f(x) = f(k)$
Hence, the function $f(x) = x-5$ is continuous at every real number
Question 3(b): Examine the following functions for continuity.
$f (x) = \frac{1}{x-5} , x \neq 5$
Answer:
The given function is
$f(x ) = \frac{1}{x-5}$
For every real number k, $k \neq 5$
We get,
$f(k) = \frac{1}{k-5}\\$
$ \lim\limits_{x\rightarrow k}f(x ) = \frac{1}{k-5}\\$
$ \lim\limits_{x\rightarrow k}f(x ) = f(k)$
Hence, function $f(x ) = \frac{1}{x-5}$ continuous for every real value of $x$, $x \neq 5$
Question 3(c): Examine the following functions for continuity.
$f (x) = \frac{x ^2-25}{x+5}, x \neq -5$
Answer:
The given function is
$f(x ) = \frac{x^2-25}{x+5}$
For every real number k, $k \neq -5$
We get,
$f(k) = \frac{k^2-5^2}{k+5}= \frac{(k +5)(k-5)}{k+5} = k-5\\$
$ \lim\limits_{x\rightarrow k}f(x ) = \frac{k^2-5^2}{k+5}$
$= \frac{(k +5)(k-5)}{k+5} = k-5\\$
$ \lim\limits_{x\rightarrow k}f(x ) = f(k)$
Hence, function $f(x ) = \frac{x^2-25}{x+5}$ continuous for every real value of x , $x \neq -5$
Question 3(d): Examine the following functions for continuity. $f (x) = | x - 5|$
Answer:
The given function is
$f (x) = | x - 5|$
for $x > 5 , f(x) = x – 5$
for $x < 5 , f(x) = 5 – x$
So, there are different cases.
case(i) $x > 5$
for every real number $k > 5$ , $f(x) = x – 5$ is defined
$f(k) = k - 5\\$
$ \lim\limits_{x\rightarrow k }f(x) = k -5\\$
$ \lim\limits_{x\rightarrow k }f(x) = f(k)$
Hence, function $f(x) = x – 5$ is continuous for $x > 5$
case (ii) $x < 5$
for every real number $k < 5$ , $f(x) = 5 – x$ is defined
$f(k) = 5-k\\ \lim\limits_{x\rightarrow k }f(x) = 5 -k\\ \lim\limits_{x\rightarrow k }f(x) = f(k)$
Hence, function $f(x) = 5 – x$ is continuous for $x < 5$
case(iii) $x = 5$
for $x = 5$ , $f(x) = x – 5$ is defined
$f(5) = 5 - 5=0\\ \lim\limits_{x\rightarrow 5 }f(x) = 5 -5=0\\ \lim\limits_{x\rightarrow 5 }f(x) = f(5)$
Hence, function $f(x) = x – 5$ is continous for $x = 5$
Hence, the function $f (x) = | x - 5|$ is continuous for every real number.
Question 4: Prove that the function $f (x) = x^n$ is continuous at x = n, where n is a positive integer
Answer:
The given function is
$f (x) = x^n$
The function $f (x) = x^n$ is defined for all positive integer, n
$f(n) = n^n\\$
$\lim\limits_{x\rightarrow n}f(x) = n^n\\$
$ \lim\limits_{x\rightarrow n}f(x) = f(n)$
Hence, the function $f (x) = x^n$ is continuous at x = n, where n is a positive integer
Question 5: Is the function f defined by
$f (x) = \left\{\begin{matrix} x , & if x \leq 1 \\ 5 & if x \geq 1 \end{matrix}\right.$
continuous at x = 0? At x = 1? At x = 2?
Answer:
The given function is
$f (x) = \left\{\begin{matrix} x , & if x \leq 1 \\ 5 & if x \geq 1 \end{matrix}\right.$
The function is defined at $x = 0$ and its value is $0$
$f(0) = 0\\$
$ \lim\limits_{x\rightarrow 0}f(x) = f(x) = 0\\$
$ \lim\limits_{x\rightarrow 0}f(x) = f(0)$
Hence, the given function is continuous at $x = 0$
The given function is defined for $x = 1$
Now, for $x = 1$ Right-hand limit and left-hand limit are not equal.
$f(1) = 1\\ \lim\limits_{x\rightarrow 1^-}f(x) = f(x) = 1\\$
$ \lim\limits_{x\rightarrow 1^+}f(x) =f(5) = 5$
R.H.L $\neq$ L.H.L.
Therefore, the given function is not continuous at $x =1$
Given function is defined for $x = 2$ and its value at $x = 2$ is $5$
$f(2) = 2\\ \lim\limits_{x\rightarrow 2}f(x) = f(5) = 5\\\lim\limits_{x\rightarrow 2}f(x) = f(2)$
Hence, the given function is continuous at $x = 2$
Question 6: Find all points of discontinuity of f, where f is defined by
$f (x) = \left\{\begin{matrix} 2x+3 & if x \leq 2 \\ 2x-3 & if x \geq 2 \end{matrix}\right.$
Answer:
The given function is
$f (x) = \left\{\begin{matrix} 2x+3 & if x \leq 2 \\ 2x-3 & if x \geq 2 \end{matrix}\right.$
The given function is defined for every real number k
There are different cases for the given function.
case(i) $k > 2$
$f(k) = 2k-3\\ \lim\limits_{x\rightarrow k}f(x) = 2k-3\\$
$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k > 2$
case(ii) $k < 2$
$f(k) = 2k +3\\ \lim\limits_{x\rightarrow k}f(x) = 2k+3\\$
$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k < 2$
case(iii) $x = 2$
$\lim\limits_{x\rightarrow 2^-}f(x) = 2x+3 = 2\times 2 + 3 = 4 + 3 = 7\\ $
$\lim\limits_{x\rightarrow 2^+}f(x) = 2x-3 = 2\times 2-3 = 4-3 = 1$
Right hand limit at x= 2 $\neq$ Left hand limit at x = 2
Therefore, x = 2 is the point of discontinuity
Question 7: Find all points of discontinuity of f, where f is defined by
$f (x) = \left\{\begin{matrix} |x|+3 & if \: \: x \leq -3 & \\ -2x & if \: \: -3 <x< 3 & \\ 6x +2 & if \: \: x \geq 3 & \end{matrix}\right.$
Answer:
The given function is
$f (x) = \left\{\begin{matrix} |x|+3 & if \: \: x \leq -3 & \\ -2x & if \: \: -3 <x< 3 & \\ 6x +2 & if \: \: x \geq 3 & \end{matrix}\right.$
The given function is defined for every real number k
There are different cases.
case (i) $k < -3$
$f(k) = -k + 3\\ \lim\limits_{x\rightarrow k}f(x) = -k + 3\\ $
$\lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for every value of k < -3
case(ii) $k = -3$
$f(-3) = -(-3) + 3 = 6\\ \lim\limits_{x\rightarrow -3^-}f(x) = -k + 3=-(-3)+3 = 6\\$
$ \lim\limits_{x\rightarrow -3^+}f(x) = -2x = -2(-3) = 6\\$
$ R.H.L. = L.H.L. = f(-3)$
Hence, the given function is continuous for x = -3
case(iii) $-3 < k < 3$
$f(k) = -2k \\ \lim\limits_{x\rightarrow k}f(x) = -2k\\$
$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of k in -3 < k < 3 given function is continuous.
case(iv) $k = 3$
$f(3) = 6x+2 = 6\times3+2 =18+2=20\\$
$ \lim\limits_{x\rightarrow 3^-}f(x) = -2x = -2(3) = -6\\$
$ \lim\limits_{x\rightarrow 3^+}f(x) = 6x+2 = 6\times3+2 = 20\\$
$ R.H.L. = f(3) \neq L.H.L.$
Hence, $x = 3$ is the point of discontinuity
case(v) $k > 3$
$f(k) = 6k+2 \\$
$ \lim\limits_{x\rightarrow k}f(x) = 6k+2 \\$
$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for every value of k > 3
Question 8: Find all points of discontinuity of f, where f is defined by
$f (x )= \left\{\begin{matrix} \frac{|x|}{x} & if \: \: x \neq 0 \\ 0 & if \: \: x = 0 \end{matrix}\right.$
Answer:
The given function is
$f (x ) \left\{\begin{matrix} \frac{|x|}{x} & if \: \: x \neq 0 \\ 0 & if \: \: x = 0 \end{matrix}\right.$
if x > 0 , $f(x)=\frac{x}{x} = 1$
if x < 0 , $f(x)=\frac{-(x)}{x} = -1$
The given function is defined for every real number k
Now,
case(i) k < 0
$f(k) = -1\\ \lim\limits_{x\rightarrow k }f(x) = -1\\$
$ \lim\limits_{x\rightarrow k }f(x) = f(k)$
Hence, the given function is continuous for every value of k < 0
case(ii) k > 0
$f(k) = 1\\ \lim\limits_{x\rightarrow k }f(x) = 1\\$
$ \lim\limits_{x\rightarrow k }f(x) = f(k)$
Hence, the given function is continuous for every value of k > 0
case(iii) x = 0
$f(0) = 0\\ \lim\limits_{x\rightarrow 0^- }f(x) = -1\\$
$ \lim\limits_{x\rightarrow 0^+}f(x) = 1\\ f(0) \neq R.H.L. \neq L.H.L.$
Hence, 0 is the only point of discontinuity
Question 9: Find all points of discontinuity of f, where f is defined by
$f (x) = \left\{\begin{matrix} \frac{x }{|x|} & if \: \: x < 0\\ -1 & if x \geq 0 \end{matrix}\right.$
Answer:
The given function is
$f (x) = \left\{\begin{matrix} \frac{x }{|x|} & if \: \: x < 0\\ -1 & if x \geq 0 \end{matrix}\right.$
if x < 0 , $f (x) =\frac{x }{|x|} = \frac{x}{-(x)} = -1$
Now, for any value of x, the value of our function is -1
Therefore, the given function is continuous for every value of x
Hence, no point in discontinuity
Question 10: Find all points of discontinuity of f, where f is defined by
$f (x) = \left\{\begin{matrix} x+1 & if \: \: x \geq 1 \\ x^2 +1 & if x \: \: <1 \end{matrix}\right. $
Answer:
The given function is
$f (x) = \left\{\begin{matrix} x+1 & if \: \: x \geq 1 \\ x^2 +1 & if x \: \: <1 \end{matrix}\right.$
The given function is defined for every real number $k$
There are different cases for the given function.
case(i) $k > 1$
$f(k) = k+1\\ \lim\limits_{x\rightarrow k}f(x) = k+1\\$
$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k > 1$
case(ii) $k < 1$
$f(k) = k^2 +1\\ \lim\limits_{x\rightarrow k}f(x) = k^2+1\\ $
$\lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k < 1$
case(iii) $x = 1$
$\lim\limits_{x\rightarrow 1^-}f(x) = x^2+1 = 1^2 + 1 = 1 + 1 = 2\\$
$ \lim\limits_{x\rightarrow 1^+}f(x) = x+1 = 1+1 = 2\\ f(1) = 1^2+1 = 2 \\ $
$R.H.L. = L.H.L. = f(1)$
Hence, at x = 2 given function is continuous.
Therefore, no point of discontinuity
Question 11: Find all points of discontinuity of f, where f is defined by
$f ( x) = \left\{\begin{matrix} x^3 -3 & if \: \: x \leq 2\\ x ^2 +1 & if \: \: x > 2 \end{matrix}\right.$
Answer:
The given function is
$f ( x) = \left\{\begin{matrix} x^3 -3 & if \: \: x \leq 2\\ x ^2 +1 & if \: \: x > 2 \end{matrix}\right.$
The given function is defined for every real number $k$
There are different cases for the given function.
case(i) $k > 2$
$f(k) = k^2+1\\ \lim\limits_{x\rightarrow k}f(x) = k^2+1\\$
$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k > 2$
case(ii) $k < 2$
$f(k) = k^3 -3\\ \lim\limits_{x\rightarrow k}f(x) = k^3-3\\$
$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k < 2$
case(iii) $x = 2$
$\lim\limits_{x\rightarrow 2^-}f(x) = x^3-3 = 2^3- 3 = 8- 3 = 5\\$
$ \lim\limits_{x\rightarrow 2^+}f(x) = x^2+1= 2^2+1 = 4+1 = 5\\$
$ f(2) = 2^3-3 = 8 - 3 = 5\\ f(2)=R.H.L.=L.H.L.$
Hence, the given function is continuous at $x = 2$
There is no point of discontinuity
Question 12: Find all points of discontinuity of f, where f is defined by
$f (x) = \left\{\begin{matrix} x ^{10} -1 & if x \leq 1 \\ x ^2 & x > 1 \end{matrix}\right.$
Answer:
The given function is
$f (x) = \left\{\begin{matrix} x ^{10} -1 & if x \leq 1 \\ x ^2 & x > 1 \end{matrix}\right.$
The given function is defined for every real number k
There are different cases for the given function.
case(i) $k > 1$
$f(k) = k^2\\ \lim\limits_{x\rightarrow k}f(x) = k^2\\$
$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of k > 1
case(ii) $k < 1$
$f(k) = k^{10} -1\\$
$ \lim\limits_{x\rightarrow k}f(x) = k^{10}-1\\$
$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of k < 1
case(iii) x = 1
$\lim\limits_{x\rightarrow 1^-}f(x) = x^{10}-1 = 1^{10} - 1 = 1 - 1 = 0\\$
$ \lim\limits_{x\rightarrow 1^+}f(x) = x^2 = 1^2 = 1\\$
$ f(1) = x^{10}-1 = 0\ f(1) = L.H.L. \neq R.H.L.$
Hence, x = 1 is the point of discontinuity.
Question 13: Is the function defined by
$f (x) = \left\{\begin{matrix} x+5 & if x \leq 1\\ x-5 & if x > 1 \end{matrix}\right.$
A continuous function?
Answer:
The given function is
$f (x) = \left\{\begin{matrix} x+5 & if x \leq 1\\ x-5 & if x > 1 \end{matrix}\right.$
The given function is defined for every real number $k$
There are different cases for the given function.
case(i) $k > 1$
$f(k) = k-5\\ \lim\limits_{x\rightarrow k}f(x) = k-5\\$
$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k > 1$
case(ii) $k < 1$
$f(k) = k+5\\ \lim\limits_{x\rightarrow k}f(x) = k+5\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k < 1$
case(iii) $x = 1$
$\lim\limits_{x\rightarrow 1^-}f(x) = x+5 = 1 + 5 = 1 + 5 = 6\\ \lim\limits_{x\rightarrow 1^+}f(x) = x-5 = 1-5 = -4\\ f(1) = x+5 =1+5= 6 \\ L.H.L. = f(1) \neq R.H.S.$
Hence, $x = 1$ is the point of discontinuity.
Question 14: Discuss the continuity of the function f, where f is defined by
$f (x)\left\{\begin{matrix} 3 & if 0 \leq x \leq 1 \\ 4& if 1 < x < 3 \\ 5& if 3 \leq x \leq 10 \end{matrix}\right.$
Answer:
The given function is
$f (x)\left\{\begin{matrix} 3 & if 0 \leq x \leq 1 \\ 4& if 1 < x < 3 \\ 5& if 3 \leq x \leq 10 \end{matrix}\right.$
The given function is defined for every real number k
Different cases are there
case (i) $k < 1$
$f(k) = 3\\ \lim\limits_{x\rightarrow k}f(x) = 3\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for every value of $k < 1$
case(ii) $k = 1$
$f(1) = 3 \\ \lim\limits_{x\rightarrow 1^-}f(x) = 3\\ \lim\limits_{x\rightarrow 1^+}f(x) = 4\\ R.H.L. \neq L.H.L. = f(1)$
Hence, the given function is discontinuous at $x = 1$
Therefore, $x = 1$ is the point of discontinuity.
case(iii) $1 < k < 3$
$f(k) = 4 \\ \lim\limits_{x\rightarrow k}f(x) = 4\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of $k$ in $1 < k < 3$ given function is continuous.
case(iv) $k = 3$
$f(3) =5\\ \lim\limits_{x\rightarrow 3^-}f(x) = 4\\ \lim\limits_{x\rightarrow 3^+}f(x) =5\\ R.H.L. = f(3) \neq L.H.L.$
Hence, $x = 3$ is the point of discontinuity
case(v) $k > 3$
$f(k) = 5 \\ \lim\limits_{x\rightarrow k}f(x) = 5 \\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for every value of $k > 3$
case(vi) when $k < 3$
$f(k) = 4 \\ \lim\limits_{x\rightarrow k}f(x) = 4\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of $k$ in $k < 3$ given function is continuous
Question 15: Discuss the continuity of the function f, where f is defined by $f(x)\left\{\begin{matrix} 2x & if &x<0 \\ 0& if &0\leq x\leq 1 \\ 4x&if & x>1 \end{matrix}\right.$
Answer:
Given function is satisfied for all real values of $x$
case (i) $k < 0$
Hence, the function is continuous for all values of $x < 0$
case (ii) $x = 0$
L.H.L at $x= 0$
R.H.L. at $x = 0$
L.H.L. = R.H.L. = $f(0)$
Hence, the function is continuous at $x = 0$
case (iii) $k > 0$
L.H.L. = R.H.L. = $f(0)$
Hence, the function is continuous for all values of x > 0
case (iv) k < 1
Hence, the function is continuous for all values of x < 1
case (v) k > 1
Hence, the function is continuous for all values of x > 1
case (vi) x = 1
Hence, the function is not continuous at x = 1
Question 16: Discuss the continuity of the function f, where f is defined by
$f ( x ) = \left\{\begin{matrix} -2 & if x \leq -1 \\ 2x & if -1< x \leq 1 \\ 2 & if x > 1 \end{matrix}\right.$
Answer:
The given function is
$f ( x ) = \left\{\begin{matrix} -2 & if x \leq -1 \\ 2x & if -1< x \leq 1 \\ 2 & if x > 1 \end{matrix}\right.$
The given function is defined for every real number $k$
Different cases are there
case (i) $k < -1$
$f(k) = -2\\ \lim\limits_{x\rightarrow k}f(x) = -2\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for every value of $k < -1$
case(ii) k = -1
$f(-1) = -2 \\ \lim\limits_{x\rightarrow -1^-}f(x) = -2\\ \lim\limits_{x\rightarrow -1^+}f(x) = 2x = 2(-1) = -2\\ R.H.L. =L.H.L. = f(-1)$
Hence, the given function is continuous at $x = -1$
case(iii) $k > -1$
$f(k) = 2k \\ \lim\limits_{x\rightarrow k}f(x) = 2k\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for all values of $x > -1$
case(vi) $-1 < k < 1$
$f(k) = 2k \\ \lim\limits_{x\rightarrow k}f(x) = 2k\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of $k$ in $-1 < k < 1$ given function is continuous.
case(v) $k = 1$
$f(1) =2x = 2(1)=2\\ \lim\limits_{x\rightarrow 1^-}f(x) = 2x=2(1)=2\\ \lim\limits_{x\rightarrow 1^+}f(x) =2\\ R.H.L. = f(1) = L.H.L.$
Hence, at x =1 function is continuous
case(vi) $k > 1$
$f(k) = 2 \\ \lim\limits_{x\rightarrow k}f(x) = 2 \\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for every value of k > 1
case(vii) when $k < 1$
$f(k) = 2k \\ \lim\limits_{x\rightarrow k}f(x) = 2k\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of k in $k < 1$ given function is continuous.
Therefore, continuous at all points
Question 17: Find the relationship between a and b so that the function f is defined by
$f (x) = \left\{\begin{matrix} ax +1 , &if x < 3 \\ bx +3 & if x > 3 \end{matrix}\right.$
is continuous at x = 3.
Answer:
The given function is
$f (x) = \left\{\begin{matrix} ax +1 , &if x < 3 \\ bx +3 & if x > 3 \end{matrix}\right.$
For the function to be continuous at $x = 3$, R.H.L. must be equal to L.H.L.
$\lim\limits_{x\rightarrow 3^-}f(x)= ax + 1 = 3a+1\\ \lim\limits_{x\rightarrow 3^+}f(x) = bx+3=3b+3$
For the function to be continuous
$\lim\limits_{x\rightarrow 3^-}f(x) = \lim\limits_{x\rightarrow 3^+}f(x) \\ 3a + 1= 3b+3\\ 3(a-b)=2\\ a-b = \frac{2}{3}\\ a = b+\frac{2}{3}$
Question 18: For what value of l is the function defined by
$f (x) = \left\{\begin{matrix} \lambda (x^2 -2x) & if x \leq 0 \\ 4x+1 & if x > 0 \end{matrix}\right.$
continuous at x = 0? What about continuity at x = 1?
Answer:
Given function is
$f (x) = \left\{\begin{matrix} \lambda (x^2 -2x) & if x \leq 0 \\ 4x+1 & if x > 0 \end{matrix}\right.$
For the function to be continuous at $x = 0$, R.H.L. must be equal to L.H.L.
$\lim\limits_{x\rightarrow 0^-}f(x) = \lambda (x^2-2x) = 0\\ \lim\limits_{x\rightarrow 0^+}f(x) = 4x+1=1$
For the function to be continuous
$\lim\limits_{x\rightarrow 0^-}f(x) = \lim\limits_{x\rightarrow 0^+}f(x) \\ 0\neq 1$
Hence, if no value of the function is continuous at $x = 0$
For $x = 1$
$f(1)=4x+1=4(1)+1=5\\ \lim\limits_{x\rightarrow 1}f(x) =4+1=5 \\\ \lim\limits_{x\rightarrow 1}f(x) = f(x)$
Hence, the given function is continuous at $x =1$
Question 19: Show that the function defined by $g (x) = x- [x]$ is discontinuous at all integral points. Here, [x] denotes the greatest integer less than or equal to x.
Answer:
Given function is
$g (x) = x- [x]$
Given is defined for all real numbers k
$\lim\limits_{x\rightarrow k^-}f(x) = k - (k-1) = k-k+1 =1\\ \lim\limits_{x\rightarrow k^+}f(x) = k – k = 0\\ \lim\limits_{x\rightarrow k^-}f(x) \neq \lim\limits_{x\rightarrow k^+}f(x)$
Hence, by this, we can say that the function defined by $g (x) = x- [x]$ is discontinuous at all integral points
Question 20: Is the function defined by $f (x) = x^2 - sin x + 5$ continuous at x = $\pi$ ?
Answer:
Given function is
$f (x) = x^2 - sin x + 5$
Clearly, the Given function is defined at x = $\pi$
$f(\pi) = \pi^2-\sin \pi+5 =\pi^2-0+5 = \pi^2+5\\ \lim\limits_{x\rightarrow \pi}f(x) = \pi^2-\sin \pi+5 =\pi^2-0+5 = \pi^2+5\\ \lim\limits_{x\rightarrow \pi}f(x) = f(\pi)$
Hence, the function defined by $f (x) = x^2 - sin x + 5$ continuous at x = $\pi$
Question 21: Discuss the continuity of the following functions:
a) $f (x) = \sin x + \cos x$
Answer:
Given function is
$f (x) = \sin x + \cos x$
The given function is defined for all real numbers.
We, know that if two function $g(x)$ and $h(x)$ are continuous then $g(x)+h(x)$ , $g(x)-h(x)$ , $g(x).h(x)$ all are continuous
Lets take $g(x) = \sin x$ and $h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim\limits_{x\rightarrow c}g(x)$
$ = \lim\limits_{x\rightarrow c}\sin x = \lim\limits_{h\rightarrow 0}\sin (c+h)$
$ \text{We know that } \sin(a+b) = \sin a \cos b + \cos a\sin b$
$ \lim\limits_{h\rightarrow 0}\sin (c+h) =$ $\lim\limits_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h)$
$= \lim\limits_{h\rightarrow 0}\sin c\cos h + \lim\limits_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim\limits_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
$h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim\limits_{x\rightarrow c}h(x) = \lim\limits_{x\rightarrow c}\cos x = \lim\limits_{h\rightarrow 0}\cos (c+h)$
$\text{ We know that }\cos(a+b) = \cos a \cos b + \sin a\sin b$
$\lim\limits_{h\rightarrow 0}\cos (c+h) = \lim\limits_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h)$
$= \lim\limits_{h\rightarrow 0}\cos c\cos h + \lim\limits_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous function
We proved independently that $\sin x$ and $\cos x$ are continuous functions.
So, we can say that
$f(x) = g(x) + h(x) = \sin x + \cos x$ is also a continuous function
Question 21(b):Discuss the continuity of the following functions:
$f (x) = \sin x - \cos x$
Answer:
Given function is
$f (x) = \sin x - \cos x$
The given function is defined for all real numbers.
We know that if two functions g(x) and h(x) are continuous then $g(x)+h(x), g(x)-h(x), g(x).h(x)$ all are continuous
Lets take $g(x) = \sin x$ and $h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim\limits_{x\rightarrow c}g(x) = \lim\limits_{x\rightarrow c}\sin x = \lim\limits_{h\rightarrow 0}\sin (c+h)$
$\text{We know that} \sin(a+b) = \sin a \cos b + \cos a\sin b$
$\lim\limits_{h\rightarrow 0}\sin (c+h) = \lim\limits_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h)$
$= \lim\limits_{h\rightarrow 0}\sin c\cos h + \lim\limits_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim\limits_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
$h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim\limits_{x\rightarrow c}h(x) = \lim\limits_{x\rightarrow c}\cos x = \lim\limits_{h\rightarrow 0}\cos (c+h)$
$ \text{We know that} \cos(a+b) = \cos a \cos b + \sin a\sin b$
$ \lim\limits_{h\rightarrow 0}\cos (c+h) = \lim\limits_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h)$
$= \lim\limits_{h\rightarrow 0}\cos c\cos h + \lim\limits_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous discussion of the continuity of the following functions:
$f (x)$ is continuous, we can say that
$f(x) = g(x) - h(x) = \sin x - \cos x$ is also a continuous function.
Question 21(c): Discuss the continuity of the following functions:
$f (x) = \sin x \cdot \cos x$
Answer:
Given function is
$f (x) = \sin x \cdot \cos x$
The given function is defined for all real numbers.
We know that if two functions $g(x)$ and $h(x)$ are continuous then $g(x)+h(x), g(x)-h(x), g(x).h(x)$ all are continuous
Lets take $g(x) = \sin x$ and $h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim\limits_{x\rightarrow c}g(x) = \lim\limits_{x\rightarrow c}\sin x = \lim\limits_{h\rightarrow 0}\sin (c+h)$
$\text{We know that} \sin(a+b) = \sin a \cos b + \cos a\sin b$
$ \lim\limits_{h\rightarrow 0}\sin (c+h) = \lim\limits_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h)$
$= \lim\limits_{h\rightarrow 0}\sin c\cos h + \lim\limits_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim\limits_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
$h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c, \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim\limits_{x\rightarrow c}h(x) = \lim\limits_{x\rightarrow c}\cos x = \lim\limits_{h\rightarrow 0}\cos (c+h)$
$\text{We know that} \cos(a+b) = \cos a \cos b + \sin a\sin b$
$ \lim\limits_{h\rightarrow 0}\cos (c+h) = \lim\limits_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h)$
$= \lim\limits_{h\rightarrow 0}\cos c\cos h + \lim\limits_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous function
We proved independently that sin x and cos x are continuous functions.
So, we can say that
$f(x) = g(x).h(x) = \sin x .\cos x$ is also a continuous function
Question 22: Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
Answer:
We know that if two functions g(x) and h(x) are continuous, then.
$\frac{g(x)}{h(x)} , h(x) \neq0\text{ is continuous}$
$\frac{1}{h(x)} , h(x) \neq 0\ \text{is continuous} \frac{1}{g(x)} , g(x) \neq0\text{ is continuous}$
Lets take $g(x) = \sin x$ and $h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim\limits_{x\rightarrow c}g(x) = \lim\limits_{x\rightarrow c}\sin x = \lim\limits_{h\rightarrow 0}\sin (c+h)$
$\text {We know that } \sin(a+b) = \sin a \cos b + \cos a\sin b$
$\lim\limits_{h\rightarrow 0}\sin (c+h) = \lim\limits_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h)$
$= \lim\limits_{h\rightarrow 0}\sin c\cos h + \lim\limits_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim\limits_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
$h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim\limits_{x\rightarrow c}h(x)$
$= \lim\limits_{x\rightarrow c}\cos x $
$= \lim\limits_{h\rightarrow 0}\cos (c+h)$
$\text{We know that }\cos(a+b) = \cos a \cos b + \sin a\sin b$
$\lim\limits_{h\rightarrow 0}\cos (c+h)$
$=\lim\limits_{h \rightarrow 0} (\cos c \cos h+ \sin c \sin h)$
$= \lim\limits_{h\rightarrow 0}\cos c\cos h + \lim\limits_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, the function $h(x) = \cos x$ is a continuous function
We proved independently that $\sin x$ and $\cos x$ a continuous functions.
So, we can say that
cosec x = $\frac{1}{\sin x} = \frac{1}{g(x)}$ is also continuous except at $x=n\pi$
sec x = $\frac{1}{\\cos x} = \frac{1}{h(x)}$ is also continuous except at $x=\frac{(2n+1) \pi}{2}$
cot x = $\frac{\cos x}{\sin x} = \frac{h(x)}{g(x)}$ is also continuous except at $x=n\pi$
Question 23: Find all points of discontinuity of f, where
$f (x ) = \left\{\begin{matrix} \frac{\sin x }{x} & if x < 0 \\ x+1 & if x > 0 \end{matrix}\right.$
Answer:
Given function is
$f (x ) = \left\{\begin{matrix} \frac{\sin x }{x} & if x < 0 \\ x+1 & if x > 0 \end{matrix}\right.$
$\lim\limits_{x\rightarrow 0^-}f(x) = \lim\limits_{x\rightarrow 0}\frac{\sin x}{x} = 1\\ \lim\limits_{x\rightarrow 0^+}f(x) = x + 1 = 1\\ \lim\limits_{x\rightarrow 0^-}f(x) = \lim\limits_{x\rightarrow 0^+}f(x)$
Hence, the function is continuous.
Therefore, no point of discontinuity
Question 24: Determine if f is defined by
$f (x) = \left\{\begin{matrix} x^2 \sin 1/x & if x \neq 0 \\ 0 & if x = 0 \end{matrix}\right.$
Is it a continuous function?
Answer:
The given function is
$f (x) = \left\{\begin{matrix} x^2 \sin 1/x & if x \neq 0 \\ 0 & if x = 0 \end{matrix}\right.$
The given function is defined for all real numbers k
when x = 0
$f(0) = 0\\ \lim\limits_{x\rightarrow 0}f(x)=\lim\limits_{x\rightarrow 0}\left ( x^2\sin\frac{1}{x} \right )=\lim\limits_{x\rightarrow 0}\left ( \frac{x.\sin\frac{1}{x}}{\frac{1}{x}} \right ) = 0(1)=0 \ \ \ \ \ \ (\because\lim\limits_{x\rightarrow 0}\frac{\sin x}{x} = 1)$
$\lim\limits_{x\rightarrow 0}f(x) = f(0)$
Hence, the function is continuous at x = 0
when $x \neq 0$
$f(k) = k^2\sin \frac{1}{k}\\ \lim\limits_{x\rightarrow k}f(x)=\lim\limits_{x\rightarrow k}\left ( x^2\sin\frac{1}{x} \right )=k^2\sin \frac{1}{k}\\ \lim\limits_{x\rightarrow k} = f(k)$
Hence, the given function is continuous for all points
Question 25: Examine the continuity of f, where f is defined by
$f (x) = \left\{\begin{matrix} \sin x - \cos x & if x \neq 0 \\ -1 & if x = 0 \end{matrix}\right.$
Answer:
The given function is
$f (x) = \sin x - \cos x$
The given function is defined for all real numbers.
We know that if two functions $g(x)$ and $h(x)$ are continuous then $g(x)+h(x), g(x)-h(x), g(x).h(x)$ all are continuous
Lets take $g(x) = \sin x$ and $h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim\limits_{x\rightarrow c}g(x) = \lim\limits_{x\rightarrow c}\sin x = \lim\limits_{h\rightarrow 0}\sin (c+h)$
$\text{We know that} \sin(a+b) = \sin a \cos b + \cos a\sin b$
$\lim\limits_{h\rightarrow 0}\sin (c+h) = \lim\limits_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h)$
$= \lim\limits_{h\rightarrow 0}\sin c\cos h + \lim\limits_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim\limits_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
$h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim\limits_{x\rightarrow c}h(x) = \lim\limits_{x\rightarrow c}\cos x$
$= \lim\limits_{h\rightarrow 0}\cos (c+h)$
$\text{We know that} \cos(a+b) = \cos a \cos b + \sin a\sin b$
$\lim\limits_{h\rightarrow 0}\cos (c+h) = \lim\limits_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h)$
$= \lim\limits_{h\rightarrow 0}\cos c\cos h + \lim\limits_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous function
We proved independently that $ \sin x$ and $\cos x$ are continuous functions.
So, we can say that
$f(x) = g(x) - h(x) =$ $\sin x$ - $\cos x$ is also a continuous function
When $x = 0$
$f (0) = (-1) \lim\limits_{x\rightarrow 0^-}f(x)$
$= \sin 0 - \cos 0 = -1$
$ \lim\limits_{x\rightarrow 0^+}f(x) = \sin 0 - \cos 0 = -1 $
$ R.H.L. = L.H.L. = f(0)$
Hence, the function is also continuous at $x = 0$
Question 26: Find the values of k so that the function f is continuous at the indicated point in Exercises
$f (x) = \left\{\begin{matrix} \frac{k \cos x }{\pi - 2x } & if x \neq \pi/2 \\ 3 & if x = \pi/2 \end{matrix}\right. \: \: \: at \: \: x = \pi /2$
Answer:
The given function is
$f (x) = \left\{\begin{matrix} \frac{k\cos x }{\pi - 2x } & if x \neq \pi/2 \\ 3 & if x = \pi/2 \end{matrix}\right.$
When $x = \frac{\pi}{2}$
$f(\frac{\pi}{2}) = 3\\let\ x=\pi +h\\ \lim\limits_{x\rightarrow \frac{\pi}{2}}f(x)= \lim\limits_{h\rightarrow 0}\frac{k\cos\left ( \frac{\pi}{2}+h \right )}{\pi-2\left ( \frac{\pi}{2}+h \right )} = k. \lim\limits_{h\rightarrow 0}\frac{-\sin h}{-2h} = \frac{k}{2}\\$
For the function to be continuous
$\lim\limits_{x\rightarrow \frac{\pi}{2}}f(x)= f(\frac{\pi}{2})\\ \frac{k}{2} = 3\\ k = 6$
Therefore, the value of k so that the function f is continuous at 6
Question 27: Find the values of k so that the function f is continuous at the indicated point in Exercises
$f (x) = \left\{\begin{matrix} kx^2 &if x \leq 2 \\ 3 & if x > 2 \end{matrix}\right. \: \: at \: \: x = 2$
Answer:
The given function is
$f (x) = \left\{\begin{matrix} kx^2 &if x \leq 2 \\ 3 & if x > 2 \end{matrix}\right.$
When $x = 2$
For the function to be continuous
$f(2) = R.H.L. = LH.L.$
$f(2) = 4k\\ \lim\limits_{x\rightarrow 2^-}f(x)= 4k\\ \lim\limits_{x\rightarrow 2^+}f(x) = 3\\ f(2) = \lim\limits_{x\rightarrow 2^-}f(x) = \lim\limits_{x\rightarrow 2^+}f(x)\\ 4k = 3\\ k = \frac{3}{4}$
Hence, the values of k so that the function f is continuous at x=2 are $\frac{3}{4}$
Question 28: Find the values of k so that the function f is continuous at the indicated point in Exercises
$f (x) = \left\{\begin{matrix} kx + 1 & if x \leq \pi \\ \cos x & if x > \pi \end{matrix}\right. \: \: at \: \: x = \pi$
Answer:
The given function is
$f (x) = \left\{\begin{matrix} kx + 1 & if x \leq \pi \\ \cos x & if x > \pi \end{matrix}\right.$
When x = $\pi$
For the function to be continuous
, f( $\pi$ ) = R.H.L. = LH.L.
$f(\pi) = k\pi+1\\ \lim\limits_{x\rightarrow \pi^-}f(x)= k\pi+1\\ \lim\limits_{x\rightarrow \pi^+}f(x) = \cos \pi = -1\\ f(\pi) = \lim\limits_{x\rightarrow \pi^-}f(x) = \lim\limits_{x\rightarrow \pi^+}f(x)\\ k\pi+1 = -1\\ k = \frac{-2}{\pi}$
Hence, the values of k so that the function f is continuous at x= $\pi$ is $\frac{-2}{\pi}$
Question 29: Find the values of k so that the function f is continuous at the indicated point in Exercises
$f (x) = \left\{\begin{matrix} kx +1 & if x \leq 5 \\ 3x-5 & if x > 5 \end{matrix}\right. \: \: at x = 5$
Answer:
Given function is
$f (x) = \left\{\begin{matrix} kx +1 & if x \leq 5 \\ 3x-5 & if x > 5 \end{matrix}\right.$
When $x = 5$
For the function to be continuous
$f(5) = R.H.L. = LH.L.$
$f(5) = 5k+1\\ \lim\limits_{x\rightarrow 5^-}f(x)= 5k+1\\ \lim\limits_{x\rightarrow 5^+}f(x) = 3(5)-5 = 15-5=10\\ f(5) = \lim\limits_{x\rightarrow 5^-}f(x) = \lim\limits_{x\rightarrow 5^+}f(x)\\ 5k+1 = 10\\ k = \frac{9}{5}$
Hence, the values of k so that the function f is continuous at $x= 5$ is $\frac{9}{5}$
Question 30: Find the values of a and b such that the function defined by
$f (x) = \left\{\begin{matrix} 5 & if\: \: x \leq 2 \\ ax + b & if\: \: 2 < x < 10 \\ 21 , & if\: \: x > 10 \end{matrix}\right.$
It is a continuous function.
Answer:
Given that continuous function is
$f (x) = \left\{\begin{matrix} 5 & if\: \: x \leq 2 \\ ax + b & if\: \: 2 < x < 10 \\ 21 , & if\: \: x > 10 \end{matrix}\right.$
The function is continuous so
$\lim\limits_{x\rightarrow 2^-}f(x) = \lim\limits_{x\rightarrow 2^+}f(x)\\ and\\ \lim\limits_{x\rightarrow 10^-}f(x)=\lim\limits_{x\rightarrow 10^+}f(x)$
$\lim\limits_{x\rightarrow 2^-}f(x) = 5\\ \lim\limits_{x\rightarrow 2^+}f(x)=ax+b=2a+b$
$ 2a+b = 5 \ \ \ \ \ \ -(i)$
$\lim\limits_{x\rightarrow 10^-}f(x)=ax+b=10a+b\\ \lim\limits_{x\rightarrow 10^+}f(x)=21$
$ 10a+b=21 \ \ \ \ \ \ \ \ -(ii)$
By solving equation (i) and (ii)
a = 2 and b = 1
Hence, values of a and b such that the function defined by $f (x) = \left\{\begin{matrix} 5 & if\: \: x \leq 2 \\ ax + b & if\: \: 2 < x < 10 \\ 21, & if\: \: x > 10 \end{matrix}\right.$ is a continuous function is 2 and 1 respectively
Question 31 Show that the function defined by $f (x) = \cos (x^2 )$ is a continuous function.
Answer:
Given function is
$f (x) = \cos (x^2 )$
Given function is defined for all real values of x
Let x = k + h
if $x\rightarrow k , \ then \ h \rightarrow 0$
$f(k) = \cos k^2\\ \lim\limits_{x \rightarrow k}f(x) = \lim\limits_{x \rightarrow k}\cos x^2 = \lim\limits_{h \rightarrow 0}\cos (k+h)^2 $
$= \cos k^2\\ \lim\limits_{x \rightarrow k}f(x) = f(k)$
Hence, the function $f (x) = \cos (x^2 )$ is a continuous function
Question 32: Show that the function defined by $f (x) = |\cos x |$ is continuous.
Answer:
Given function is
$f (x) = |\cos x |$
Given function is defined for all values of x
f = g o h , $g(x) = \|x\|$ and $h(x) = \cos x$
Now,
$g(x)\begin{cases} -x & \text{ if } x<0 \\ 0 & \text{ if } x= 0\\ x& \text{ if } x>0 \end{cases}$
g(x) is defined for all real numbers k
case(i) $k < 0$
$g(k) = -k\\ \lim\limits_{x\rightarrow k}g(x) = -k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when $k < 0$
case (ii) $k > 0$
$g(k) = k\\ \lim\limits_{x\rightarrow k}g(x) = k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k > 0
case (iii) $k = 0$
$g(0) = 0\\ \lim\limits_{x\rightarrow 0^-}g(x) = -x = 0$
$ \lim\limits_{x\rightarrow 0^+}g(x ) = x = 0$
$\lim\limits_{x\rightarrow 0^-}g(x) = g(0) = \lim\limits_{x\rightarrow 0^+}g(x )$
Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x
Now,
$h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim\limits_{x\rightarrow c}h(x)$
$= \lim\limits_{x\rightarrow c}\cos x$
$= \lim\limits_{h\rightarrow 0}\cos (c+h)$
$\text{We know that } \cos(a+b) = \cos a \cos b + \sin a\sin b$
$ \lim\limits_{h\rightarrow 0}\cos (c+h) = \lim\limits_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h)$
$= \lim\limits_{h\rightarrow 0}\cos c\cos h + \lim\limits_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous function
g(x) is continuous , $h(x)$ is continuous
Therefore, $f(x) = g o h$ is also continuous
Question 33: Examine that sin | x| is a continuous function.
Answer:
The given function is
$f(x) = \sin \|x\|$
f(x) = h o g , h(x) =$\sin x$ and g(x) = |x|
Now,
$g(x)\begin{cases} -x & \text{ if } x<0 \\ 0 & \text{ if } x= 0\\ x& \text{ if } x>0 \end{cases}$
g(x) is defined for all real numbers k
case(i) $k < 0$
$g(k) = -k\\ \lim\limits_{x\rightarrow k}g(x) = -k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k < 0
case (ii) $k > 0$
$g(k) = k\\ \lim\limits_{x\rightarrow k}g(x) = k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k > 0
case (iii) $k = 0$
$g(0) = 0\\ \lim\limits_{x\rightarrow 0^-}g(x) = -x = 0$
$\lim\limits_{x\rightarrow 0^+}g(x ) = x = 0$
$ \lim\limits_{x\rightarrow 0^-}g(x) = g(0) = \lim\limits_{x\rightarrow 0^+}g(x )$
Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x
Now,
h(x) =$\sin x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \sin c$
$ \lim\limits_{x\rightarrow c}h(x) = \lim\limits_{x\rightarrow c}\sin x$
$= \lim\limits_{h\rightarrow 0}\sin (c+h)$
$\text{ We know that } \sin(a+b) = \sin a \cos b + \cos a\sin b$
$ \lim\limits_{h\rightarrow 0}\sin (c+h) = \lim\limits_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h)$
$= \lim\limits_{h\rightarrow 0}\sin c\cos h + \lim\limits_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \sin x$ is a continuous function
g(x) is continuous , h(x) is continuous
Therefore, $f(x) = h o g$ is also continuous
Question 34: Find all the points of discontinuity of f defined by $f (x) = | x| - | x + 1|.$
Answer:
The given function is
$f (x) = | x| - | x + 1|$
Let g(x) = |x| and h(x) = |x+1|
Now,
$g(x)\begin{cases} -x & \text{ if } x<0 \\ 0 & \text{ if } x= 0\\ x& \text{ if } x>0 \end{cases}$
g(x) is defined for all real numbers k
case(i) k < 0
$g(k) = -k\\ \lim\limits_{x\rightarrow k}g(x) = -k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k < 0
case (ii) k > 0
$g(k) = k\\ \lim\limits_{x\rightarrow k}g(x) = k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k > 0
case (iii) k = 0
$g(0) = 0\\ \lim\limits_{x\rightarrow 0^-}g(x) = -x = 0$
$ \lim\limits_{x\rightarrow 0^+}g(x ) = x = 0$
$ \lim\limits_{x\rightarrow 0^-}g(x) = g(0) = \lim\limits_{x\rightarrow 0^+}g(x )$
Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x
Now,
$h(x)\begin{cases} -(x+1) & \text{ if } x<-1 \\ 0 & \text{ if } x= -1\\ (x+1)& \text{ if } x>-1 \end{cases}$
g(x) is defined for all real numbers k
case(i) k < -1
$h(k) = -(k+1)\\ \lim\limits_{x\rightarrow k}h(x) = -(k+1)\\ \lim\limits_{x\rightarrow k}h(x) = h(k)$
Hence, h(x) is continuous when k < -1
case (ii) k > -1
$h(k) = k+1\\ \lim\limits_{x\rightarrow k}h(x) = k+1\\ \lim\limits_{x\rightarrow k}h(x) = h(k)$
Hence, h(x) is continuous when k > -1
case (iii) k = -1
$h(-1) = 0\\ \lim\limits_{x\rightarrow -1^-}h(x) = -(x-1)$
$= 0\\ \lim\limits_{x\rightarrow -1^+}h(x ) = x+1 = 0$
$ \lim\limits_{x\rightarrow -1^-}h(x) = h(0) = \lim\limits_{x\rightarrow -1^+}h(x )$
Hence, h(x) is continuous when k = -1
Therefore, h(x) = |x+1| is continuous for all real values of x.
g(x) is continuous and h(x) is continuous
Therefore, $f(x) = g(x) - h(x) = |x| - |x+1|$ is also continuous
|
NCERT Continuity and Differentiability Class 12 Solutions: Exercise: 5.2
Page number: 122
Total questions: 10
|
Question 1: Differentiate the functions with respect to x in
$\sin (x^2 +5 )$
Answer:
The given function is
$f(x)=\sin (x^2 +5 )$
When we differentiate it w.r.t. x.
Let's take $t = x^2+5$. then,
$f(t) = \sin t$
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx}$ (By chain rule)
$\frac{df(t)}{dt} = \frac{d(\sin t )}{dt} = \cos t = \cos (x^2+5)$
$\frac{dt}{dx} = \frac{d(x^2+5 )}{dx} = 2x$
Now,
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = \cos (x^2+5).2x$
Therefore, the answer is $2x \cos (x^2+5)$
Question 2: Differentiate the functions with respect to x in
$\cos ( \sin x )$
Answer:
The given function is
$f(x)= \cos ( \sin x )$
Let’s take $t = \sin x$ then,
$f(t) = \cos t$
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx}$ ( By chain rule)
$\frac{df(t)}{dt} = \frac{d(\cos t)}{dt} = -\sin t = -\sin (\sin x)$
$\frac{dt}{dx} = \frac{d(\sin x)}{dt} = \cos x$
Now,
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = -\sin(\sin x).\cos x$
Therefore, the answer is $-\sin(\sin x).\cos x$
Question 3: Differentiate the functions with respect to x in
$\sin (ax +b )$
Answer:
The given function is
$f(x) = \sin (ax +b )$
When we differentiate it w.r.t. x.
Let's take $t = ax+b$. then,
$f(t) = \sin t$
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx}$ (By chain rule)
$\frac{df(t)}{dt} = \frac{d(\sin t )}{dt} = \cos t = \cos (ax+b)$
$\frac{dt}{dx} = \frac{d(ax+b )}{dx} = a$
Now,
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = \cos (ax+b).a$
Therefore, the answer is $a \cos (ax+b)$
Question 4: Differentiate the functions with respect to x in
$\sec (\tan (\sqrt x) )$
Answer:
The given function is
$f(x)=\sec (\tan (\sqrt x) )$
When we differentiate it w.r.t. x.
Let's take $t = \sqrt x$. then,
$f(t) = \sec (\tan t)$
take $\tan t = k$ . then,
$f(k) = \sec k$
$\frac{df(k)}{dx} = \frac{df(k)}{dk}.\frac{dk}{dt}.\frac{dt}{dx}$ (By chain rule)
$\frac{df(k)}{dk} = \frac{d(\sec k )}{dk} = \sec k \tan k = \sec(\tan\sqrt x)\tan(\tan\sqrt x)$
$(\because k = \tan t \ and \ t = \sqrt x)$
$\frac{df(t)}{dt} = \frac{d(\tan t )}{dt} = \sec^2 t =\sec^2 (\sqrt x) \ \ \ \ \ \ (\because t = \sqrt x)$
$\frac{dt}{dx} = \frac{d(\sqrt x)}{dx} = \frac{1}{2\sqrt x}$
Now,
$\frac{df(k)}{dx} = \frac{df(k)}{dk}.\frac{dk}{dt}.\frac{dt}{dx}$
$=\sec(\tan \sqrt x)\tan(\tan \sqrt x).\sec^2 (\sqrt x) . \frac{1}{2\sqrt x}$
Therefore, the answer is $\frac{\sec(\tan \sqrt x).\tan(\tan \sqrt x).\sec^2 (\sqrt x)}{2\sqrt x}$
Question 5: Differentiate the functions with respect to x in
$\frac{\sin (ax +b )}{\cos (cx + d)}$
Answer:
The given function is
$f(x) = \frac{\sin (ax +b )}{\cos (cx + d)} = \frac{g(x)}{h(x)}$
We know that,
$f^{'}(x) = \frac{g^{'}(x)h(x)-g(x)h^{'}(x)}{h^2(x)}$
$g(x) = \sin(ax+b)$ and $h(x) = \cos(cx+d)$
Let's take $u = (ax+b) \ and \ v = (cx+d)$
Then,
$\sin (ax+b) = \sin u \ and \ \cos(cx+d) = \cos c$
$g^{'}(x)=\frac{d(g(x))}{dx} = \frac{d(g(x))}{du}.\frac{du}{dx}$ (By chain rule)
$\frac{d(g(x))}{du} = \frac{d(\sin u)}{du} = \cos u = \cos(ax+b) \ \ \ \ \ \ \ \ \ (\because u = ax +b)$
$\frac{du}{dx} = \frac{d(ax+b)}{dx} = a$
$g^{'}(x)=a\cos (ax+b)$ -(i)
Similarly,
$h^{'}(x)=\frac{d(h(x))}{dx} = \frac{d(h(x))}{dv}.\frac{dv}{dx}$
$\frac{d(h(x))}{dv}= \frac{d(\cos v)}{dv} = -\sin v = -\sin (cx+d) \ \ \ \ \ \ \ (\because v = (cx+d))$
$\frac{dv}{dx}= \frac{d(cx+d)}{dv} = c$
$h^{'}(x)=-c\sin(cx+d)$ -(ii)
Now, put (i) and (ii) in
$f^{'}(x) = \frac{g^{'}(x)h(x)-g(x)h^{'}(x)}{h^2(x)} = \frac{a\cos(ax+b).\cos(cx+d)-\sin(ax+b).(-c.\sin(cx+d))}{\cos^2(cx+d)}$
$= \frac{a\cos(ax+b).\cos(cx+d)}{\cos^2(cx+d)}+\frac{\sin(ax+b).c.\sin(cx+d)}{\cos^2(cx+d)}$
$= a\cos(ax+b).\sec(cx+d) +c\sin(ax+b).\tan(cx+d).\sec(cx+d)$
Therefore, the answer is $a\cos(ax+b).\sec(cx+d) +c\sin(ax+b).\tan(cx+d).\sec(cx+d)$
Question 6: Differentiate the functions with respect to x in
$\cos x^3 . \sin ^ 2 ( x ^5 )$
Answer:
The given function is
$f(x)=\cos x^3 . \sin ^ 2 ( x ^5 )$
Differentiation w.r.t. x is
$f^{'}(x) = g^{'}(x).h(x) + g(x).h^{'}(x)$
$g(x) = \cos x^3 \ and \ h(x) = sin^2(x^5)$
Lets take $u = x^3 \ and \ v = x^5$
Our functions become,
$\cos x^3 = \cos u$ and $\sin^2(x^5) = \sin^2v$
Now,
$g^{'}(x) = \frac{d(g(x))}{dx} =\frac{d(g(u))}{du}.\frac{du}{dx}$ ( By chain rule)
$\frac{d(g(u))}{du} = \frac{d(\cos u)}{du} = -\sin u =- \sin x^3 \ \ \ \ (\because u = x^3)$
$\frac{du}{dx} = \frac{d(x^3)}{dx} = 3x^2$
$g^{'}(x) = -\sin x^3.3x^2$ -(i)
Similarly,
$h^{'}(x) = \frac{d(h(x))}{dx} =\frac{d(h(v))}{dv}.\frac{dv}{dx}$
$\frac{d(h(v))}{dv}= \frac{d(\sin^2v)}{dv} =2\sin v \cos v =2\sin x^5\cos x^5$
$ (\because v = x^5)$
$\frac{dv}{dx} = \frac{d(x^5)}{dx} = 5x^4$
$h^{'}(x) = 2\sin x^5\cos x^5.5x^4 = 10x^4\sin x^5\cos x^5$ -(ii)
Put (i) and (ii) in
$f^{'}(x) = g^{'}(x).h(x) + g(x).h^{'}(x) $
$= -3x^2\sin x^3.\sin^2 x^5+\cos x^3.10x^4\sin x^5 \cos x^5$
Therefore, the answer is $10x^4\sin x^5 \cos x^5.\cos x^3 -3x^2\sin x^3.\sin^2 x^5$
Question 7: Differentiate the functions with respect to x in
$2 \sqrt { \cot ( x^2 )}$
Answer:
The give function is
$f(x)=2 \sqrt { \cot ( x^2 )}$
Let's take $t = x^2$
Now, take $\cot t = k^2$
$f(k) = 2k$
Differentiation w.r.t. x
$\frac{d(f(k))}{dx} = \frac{d(f(k))}{dk}.\frac{dk}{dt}.\frac{dt}{dx}$ -(By chain rule)
$\frac{d(f(k))}{dk} = \frac{d(2k)}{dk} = 2$
$\frac{dk}{dt} = \frac{d(\sqrt{\cot t})}{dt} = \frac{1}{2\sqrt{cot t}}.(-cosec^2 t) = \frac{-cosec^2 x^2}{2\sqrt{cot x^2}}(\because t = x^2)$
$\frac{dt}{dx} = \frac{d(x^2)}{dx} = 2x$
So,
$\frac{d(f(k))}{dx} = 2.\frac{-cosec^2 x^2}{2\sqrt{cot x^2}}.2x = \frac{-2\sqrt2x}{\sin^2x^2\sqrt{\frac{2\sin x^2\cos x^2}{\sin^2x^2}} }$ ( Multiply and divide by $\sqrt 2$ and multiply and divide $\sqrt {\cot x^2}$ by $\sqrt{\sin x^2}$
$(\text{because} \cot x = \frac{\cos x}{\sin x} \ and \csc x = \frac{1}{\sin x })$
$=\frac{-2\sqrt2x}{\sin x^2\sqrt{\sin2x^2}}(\because 2\sin x\cos x=\sin2x)$
There, the answer is $\frac{-2\sqrt2x}{\sin x^2\sqrt{\sin2x^2}}$
Question 8: Differentiate the functions with respect to x in
$\cos ( \sqrt x )$
Answer:
Let us assume : $y\ =\ \cos ( \sqrt x )$
Differentiating y with respect to x, we get :
$\frac{dy}{dx}\ =\ \frac{d(\cos ( \sqrt x ))}{dx}$
or $=\ - \sin \sqrt{x}.\frac{d( \sqrt x )}{dx}$
or $=\ \frac{- \sin \sqrt{x}}{2\sqrt{x}}$
Question 9: Prove that the function f given by $f (x) = |x-1 |, x \epsilon R$ is not differentiable at x = 1.
Answer:
The given function is
$f (x) = |x-1 | , x \epsilon R$
We know that any function is differentiable when both.
$\lim\limits_{h\rightarrow 0^-}\frac{f(c+h)-f(c)}{h}$ and $\lim\limits_{h\rightarrow 0^+}\frac{f(c+h)-f(c)}{h}$ are finite and equal
The required condition for the function to be differentiable at x = 1 is
$\lim\limits_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim\limits_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h}$
Now, the Left-hand limit of a function at x = 1 is
$\lim\limits_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim\limits_{h\rightarrow 0^-}\frac{|1+h-1|-|1-1|}{h} = \lim\limits_{h\rightarrow 0^-}\frac{|h|-0}{h}$
$= \lim\limits_{h\rightarrow 0^-}\frac{-h}{h} = -1 \ \ \ \ (\because h < 0)$
The right-hand limit of a function at x = 1 is
$\lim\limits_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h } = \lim\limits_{h\rightarrow 0^+}\frac{|1+h-1|-|1-1|}{h} = \lim\limits_{h\rightarrow 0^+}\frac{|h|-0}{h}$
$=\lim\limits_{h\rightarrow 0^-}\frac{h}{h} = 1$
Now, it is clear that.
R.H.L. at x= 1 $\neq$ L.H.L. at x= 1
Therefore, function $f (x) = |x-1 |$ is not differentiable at x = 1
Question 10: Prove that the greatest integer function defined by $f (x) = [x] , 0 < x < 3$ is not differentiable at
x = 1 and x = 2.
Answer:
The given function is
$f (x) = [x] , 0 < x < 3$
We know that any function is differentiable when both.
$\lim\limits_{h\rightarrow 0^-}\frac{f(c+h)-f(c)}{h}$ and $\lim\limits_{h\rightarrow 0^+}\frac{f(c+h)-f(c)}{h}$ are finite and equal
The required condition for the function to be differentiable at x = 1 is
$\lim\limits_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim\limits_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h}$
Now, the Left-hand limit of the function at x = 1 is
$\lim\limits_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim\limits_{h\rightarrow 0^-}\frac{[1+h]-[1]}{h} = \lim\limits_{h\rightarrow 0^-}\frac{0-1}{h}$
$=\lim\limits_{h\rightarrow 0^-}\frac{-1}{h} = -\infty \ \ \ \ \ (\because h < 0 \rightarrow 1+h<1, \therefore [1+h] =0)$
The right-hand limit of the function at x = 1 is
$\lim\limits_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h } = \lim\limits_{h\rightarrow 0^+}\frac{[1+h]-[1]}{h} = \lim\limits_{h\rightarrow 0^+}\frac{1-1}{h}$
$=\lim\limits_{h\rightarrow 0^-}\frac{0}{h} = 0 \ \ \ \ \ (\because h > 0 \rightarrow 1+h>1, \therefore [1+h] =1)$
Now, it is clear that.
R.H.L. at x= 1 $\neq$ L.H.L. at x= 1 and L.H.L. is not finite as well
Therefore, function $f(x) = [x]$ is not differentiable at x = 1
Similarly, for x = 2
The required condition for the function to be differentiable at x = 2 is
$\lim\limits_{h\rightarrow 0^-}\frac{f(2+h)-f(2)}{h } = \lim\limits_{h\rightarrow 0^+}\frac{f(2+h)-f(2)}{h}$
Now, the Left-hand limit of the function at x = 2 is
$\lim\limits_{h\rightarrow 0^-}\frac{f(2+h)-f(2)}{h } = \lim\limits_{h\rightarrow 0^-}\frac{[2+h]-[2]}{h} = \lim\limits_{h\rightarrow 0^-}\frac{1-2}{h}$
$=\lim\limits_{h\rightarrow 0^-}\frac{-1}{h} = -\infty \ \ \ \ (\because h < 0 \rightarrow 2+h<2, \therefore [2+h] =1)$
The right-hand limit of the function at x = 1 is
$\lim\limits_{h\rightarrow 0^+}\frac{f(2+h)-f(2)}{h } = \lim\limits_{h\rightarrow 0^+}\frac{[2+h]-[2]}{h} = \lim\limits_{h\rightarrow 0^+}\frac{2-2}{h}$
$=\lim\limits_{h\rightarrow 0^-}\frac{0}{h} = 0 \ \ \ \ \ (\because h > 0 \rightarrow 2+h>2, \therefore [2+h] =2)$
Now, it is clear that.
R.H.L. at x= 2 $\neq$ L.H.L. at x= 2 and L.H.L. is not finite as well
Therefore, function $f(x) = [x]$ is not differentiable at x = 2
|
NCERT Continuity and Differentiability Class 12 Solutions: Exercise: 5.3
Page number: 125
Total questions: 15
|
Question 1: Find $\frac{dy}{dx}$ in the following:
$2 x + 3 y = \sin x$
Answer:
The given function is
$2 x + 3 y = \sin x$
We can rewrite it as
$3y = \sin x - 2x$
Now, differentiation w.r.t. x is
$3\frac{dy}{dx} = \frac{d(\sin x - 2x)}{dx} = \cos x - 2$
$\frac{dy}{dx} = \frac{\cos x-2}{3}$
Therefore, the answer is $\frac{\cos x-2}{3}$
Question 2: Find $\frac{dy}{dx}$ in the following: $2 x + 3y = \sin y$
Answer:
The given function is
$2 x + 3 y = \sin y$
We can rewrite it as
$\sin y - 3y = 2x$
Now, differentiation w.r.t. x is
$\frac{dy}{dx}(\sin y - 3y) = \frac{d( 2x)}{dx}$
$(\cos y\frac{dy}{dx} - 3\frac{dy}{dx}) = 2$
$\frac{dy}{dx} = \frac{2}{\cos y -3}$
Therefore, the answer is $\frac{2}{\cos y -3}$
Question 3: Find $\frac{dy}{dx}$ in the following: $ax + by ^2 = \cos y$
Answer:
The given function is
$ax + by ^2 = \cos y$
We can rewrite it as
$by^2-\cos y = -ax$
Now, differentiation w.r.t. x is
$\frac{dy}{dx}(2by - (-\sin y)) = \frac{d( -ax)}{dx} = -a$
$\frac{dy}{dx} = \frac{-a}{2b y +\sin y}$
Therefore, the answer is $\frac{-a}{2b y +\sin y}$
Question 4: Find $\frac{dy}{dx}$ in the following:
$xy + y^2 = \tan x + y$
Answer:
The given function is
$xy + y^2 = \tan x + y$
We can rewrite it as
$xy+y^2-y= \tan x$
Now, differentiation w.r.t. x is
$y+\frac{dy}{dx}(x+2y-1) = \frac{d( \tan x)}{dx} = \sec^2 x$
$\frac{dy}{dx} = \frac{\sec^2 x- y}{x+2y-1}$
Therefore, the answer is $\frac{\sec^2 x- y}{x+2y-1}$
Question 5: Find $\frac{dy}{dx}$ in the following: $x^2 + xy + y^2 = 100$
Answer:
The given function is
$x^2 + xy + y^2 = 100$
We can rewrite it as
$xy + y^2 = 100 - x^2$
Now, differentiation w.r.t. x is
$y+\frac{dy}{dx}(x+2y) = \frac{d( 100-x^2)}{dx} = -2x$
$\frac{dy}{dx} = \frac{-2 x- y}{x+2y}$
Therefore, the answer is $\frac{-2 x- y}{x+2y}$
Question 6: Find $\frac{dy}{dx}$ in the following:
$x ^3 + x^2 y + xy^2 + y^3 = 81$
Answer:
The given function is
$x ^3 + x^2 y + xy^2 + y^3 = 81$
We can rewrite it as
$x^2 y + xy^2 + y^3 = 81 - x^3$
Now, differentiation w.r.t. x is
$\frac{d(x^2 y + xy^2 + y^3)}{dx} = \frac{d(81 - x^3)}{dx}$
$2xy+y^2+\frac{dy}{dx}(x^2+2xy+3y^2) = -3x^2\\ \frac{dy}{dx}=\frac{-(3x^2+2xy+y^2)}{(x^2+2xy+3y^2}$
Therefore, the answer is $\frac{-(3x^2+2xy+y^2)}{(x^2+2xy+3y^2}$
Question 7: Find $\frac{dy}{dx}$ in the following: $\sin ^ 2 y + \cos xy = k$
Answer:
The given function is
$\sin ^ 2 y + \cos xy = k$
Now, differentiation w.r.t. x is
$\frac{d(\sin^2y+\cos xy)}{dx} = \frac{d(k)}{dx}$
$2\sin y \cos y\frac{dy}{dx}+(-\sin xy)(y+x\frac{dy}{dx})=0$
$\frac{dy}{dx}(2\sin y \cos y-x\sin xy)= y\sin xy\\ \frac{dy}{dx}$
$= \frac{y\sin xy}{2\sin y \cos y-x\sin xy}$
$= \frac{y\sin xy}{\sin 2y -x\sin xy} \ \ \ \ \ \ (\because 2\sin x\cos y = \sin 2x)$
Therefore, the answer is $\frac{y\sin xy}{\sin 2y -x\sin xy}$
Question 8: Find $\frac{dy}{dx}$ in the following:
$\sin ^2 x + \cos ^ 2 y = 1$
Answer:
The given function is
$\sin ^2 x + \cos ^ 2 y = 1$
We can rewrite it as
$\cos ^ 2 y = 1-\sin^2x$
Now, differentiation w.r.t. x is
$\frac{d(\cos^2y)}{dx} = \frac{d(1-\sin^2x)}{dx}$
$2\cos y (-\sin y)\frac{dy}{dx} = -2\sin x \cos x\\ \frac{dy}{dx} = \frac{2\sin x\cos x}{2\sin y \cos y} = \frac{\sin 2x }{\sin 2y} \ \ \ \ \ \ (\because2\sin a \cos a = \sin 2a)$
Therefore, the answer is $\frac{\sin 2x}{\sin 2y }$
Question 9: Find $\frac{dy}{dx}$ in the following:
$y = \sin ^{-1} \left ( \frac{2x}{1+ x^2 } \right )$
Answer:
The given function is
$y = \sin ^{-1} \left ( \frac{2x}{1+ x^2 } \right )$
Lets consider $x = \tan t$
Then,
$\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)$
$1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)$
Now,
$\frac{2x}{1+x^2} = \frac{2\tan t }{1+\tan^2t} = \sin 2t \ \ \ \ \ \ (\because \sin 2x = \frac{2\tan x }{1+\tan^2 x})$
Our equation reduces to
$y = \sin^{-1}(\sin 2t)$
$y = 2t$
Now, differentiation w.r.t. x is
$\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx} = 2.\frac{1}{1+x^2} = \frac{2}{1+x^2}$
Therefore, the answer is $\frac{2}{1+x^2}$
Question 10: Find $\frac{dy}{dx}$ in the following:
$y = \tan ^{-1} \left ( \frac{3x- x ^3}{1- 3x ^2} \right ) , - \frac{1}{\sqrt 3 } < x < \frac{1}{\sqrt 3 }$
Answer:
The given function is
$y = \tan ^{-1} \left ( \frac{3x- x ^3}{1- 3x ^2} \right )$
Lets consider $x = \tan t$
Then,
$\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)$
$1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)$
Now,
$\frac{3x-x^3}{1-3x^2} = \frac{3\tan t-\tan^3t }{1-3\tan^2t} = \tan3t \ \ \ \ \ \ (\because \tan 3x = \frac{3\tan x-\tan^3x }{1-3\tan^2x} )$
Our equation reduces to
$y = \tan^{-1}(\tan 3t)$
$y = 3t$
Now, differentiation w.r.t. x is
$\frac{d(y)}{dx} = \frac{d(3t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx} = 3.\frac{1}{1+x^2} = \frac{3}{1+x^2}$
Therefore, the answer is $\frac{3}{1+x^2}$
Question 11: Find $\frac{dy}{dx}$ in the following:
$y = \cos ^{-1} \left ( \frac{1 - x^2 }{1+ x^2 } \right ) , 0 < x < 1$
Answer:
The given function is
$y = \cos ^{-1} \left ( \frac{1 - x^2 }{1+ x^2 } \right )$
Let's consider $x = \tan t$
Then,
$\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)$
$1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)$
Now,
$\frac{1-x^2}{1+x^2} = \frac{1-\tan^2t }{1+\tan^2t} = \cos 2t \ \ \ \ \ \ (\because \cos 2x = \frac{1-\tan^2x }{1+\tan^2x} )$
Our equation reduces to
$y = \cos^{-1}(\cos 2t)$
$y = 2t$
Now, differentiation w.r.t. x is
$\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx} = 2.\frac{1}{1+x^2} = \frac{2}{1+x^2}$
Therefore, the answer is $\frac{2}{1+x^2}$
Question 12: Find $\frac{dy}{dx}$ in the following: $y = \sin ^{-1 } \left ( \frac{1- x ^2 }{1+ x^2} \right ) , 0< x < 1$
Answer:
The given function is
$y = \sin ^{-1} \left ( \frac{1 - x^2 }{1+ x^2 } \right )$
We can rewrite it as
$\sin y = \ \left ( \frac{1 - x^2 }{1+ x^2 } \right )$
Let's consider $x = \tan t$
Then,
$\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)$
$1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)$
Now,
$\frac{1-x^2}{1+x^2} = \frac{1-\tan^2t }{1+\tan^2t} = \cos 2t \ \ \ \ \ \ (\because \cos 2x = \frac{1-\tan^2x }{1+\tan^2x} )$
Our equation reduces to
$\sin y = \cos 2t$
Now, differentiation w.r.t. x is
$\frac{d(\sin y)}{dx} = \frac{d(\cos2t)}{dt}.\frac{dt}{dx}$
$\cos y\frac{dy}{dx} = 2(-\sin 2t).\frac{1}{1+x^2} = \frac{-2\sin2t}{1+x^2}$ $= \frac{-2.\frac{2\tan t}{1+\tan^2t}}{1+x^2} =\frac{-2.\frac{2x}{1+x^2}}{1+x^2} =\frac{-4x}{(1+x^2)^2}$
$(\because \sin 2x = \frac{2\tan x}{1+\tan^2x} \ and \ x = \tan t)$
$\sin y = \ \left ( \frac{1 - x^2 }{1+ x^2 } \right )\rightarrow \cos y = \frac{2x}{1+x^2}$
$\frac{2x}{1+x^2}\frac{dy}{dx} = \frac{-4x}{(1+x^2)^2}$
$\frac{dy}{dx} = \frac{-2}{(1+x^2)}$
Therefore, the answer is $\frac{-2}{1+x^2}$
Question 13: Find $\frac{dy}{dx}$ in the following:
$y = \cos ^{-1} \left ( \frac{2x}{1+ x^2 } \right ) , -1 < x < 1$
Answer:
The given function is
$y = \cos ^{-1} \left ( \frac{2x}{1+ x^2 } \right )$
We can rewrite it as
$\cos y = \left ( \frac{2x}{1+ x^2 } \right )$
Let's consider $x = \tan t$
Then,
$\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)$
$1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)$
Now,
$\frac{2x}{1+x^2} = \frac{2\tan t }{1+\tan^2t} = \sin 2t \ \ \ \ \ \ (\because \sin 2x = \frac{2\tan x }{1+\tan^2x} )$
Our equation reduces to
$\cos y = \sin 2t$
Now, differentiation w.r.t. x is
$\frac{d(\cos y)}{dx} = \frac{d(\sin2t)}{dt}.\frac{dt}{dx}$
$(-\sin y)\frac{dy}{dx} = 2(\cos 2t).\frac{1}{1+x^2} = \frac{2\cos2t}{1+x^2}$ $= \frac{2.\frac{1-\tan^2 t}{1+\tan^2t}}{1+x^2} =\frac{2.\frac{1-x^2}{1+x^2}}{1+x^2} =\frac{2(1-x^2)}{(1+x^2)^2}$
$(\because \cos 2x = \frac{1-\tan^2 x}{1+\tan^2x} \ and \ x = \tan t)$
$\cos y = \ \left ( \frac{2 x }{1+ x^2 } \right )\rightarrow \sin y = \frac{1-x^2}{1+x^2}$
$-\frac{1-x^2}{1+x^2}\frac{dy}{dx} = \frac{2(1-x^2)}{(1+x^2)^2}$
$\frac{dy}{dx} = \frac{-2}{(1+x^2)}$
Therefore, the answer is $\frac{-2}{1+x^2}$
Question 14: Find $\frac{dy}{dx}$ in the following:
$y = \sin ^ { -1 } ( 2x \sqrt {1- x^2} ) , -\frac{1}{\sqrt2} < x \frac{1}{\sqrt 2 }$
Answer:
The given function is
$y = \sin ^ { -1 } ( 2x \sqrt {1- x^2} )$
Let's take $x = \sin t$
Then,
$\frac{d(x)}{dx} = \frac{(\sin t)}{dt}.\frac{dt}{dx} \ \ \ \ \ (by \ chain \ rule)$
$1 =\cos t.\frac{dt}{dx}$
$\frac{dt}{dx} = \frac{1}{\cos t } = \frac{1}{\sqrt{1-\sin ^2t}} = \frac{1}{\sqrt{1-x^2}}$
$(\because \cos x = \sqrt{1-\sin^2x} \ and \ x = \sin t )$
And
$2x\sqrt{1-x^2} = 2\sin t \sqrt{1-\sin^2t} = 2\sin t \sqrt{\cos^2 t}$
$= 2\sin t\cos t =\sin 2t$
$(\because \cos x = \sqrt{1-\sin^2x} \text{ and }\ 2\sin x\cos x = \sin2x )$
Now, our equation reduces to
$y = \sin ^ { -1 } ( \sin 2t )$
$y = 2t$
Now, differentiation with respect to. x
$\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx} = 2.\frac{1}{\sqrt{1-x^2}} = \frac{2}{\sqrt{1-x^2}}$
Therefore, the answer is $\frac{2}{\sqrt{1-x^2}}$
Question 15: Find $\frac{dy}{dx}$ in the following:
$y = \sec ^{-1} \left ( \frac{1}{2x ^2 -1} \right ) , 0 < x < 1/ \sqrt 2$
Answer:
The given function is
$y = \sec ^{-1} \left ( \frac{1}{2x ^2 -1} \right )$
Let's take $x = \cos t$
Then,
$\frac{d(x)}{dx} = \frac{(\cos t)}{dt}.\frac{dt}{dx} \ \ \ \ \ (by \ chain \ rule)$
$1 =-\sin t.\frac{dt}{dx}$
$\frac{dt}{dx} = \frac{-1}{\sin t } = \frac{-1}{\sqrt{1-\cos ^2t}} = \frac{-1}{\sqrt{1-x^2}}$
$(\because \sin x = \sqrt{1-\cos^2x} \ and \ x = \cos t )$
And
$\frac{1}{2x^2-1} =\frac{1}{2\cos^2 t - 1} = \frac{1}{\cos2t} = \sec2t$
$(\because \cos 2x = \sqrt{2\cos^2x-1} \text{ and }\cos x = \frac{1}{\sec x} )$
Now, our equation reduces to
$y = \sec ^{-1} \left ( \sec 2t \right )$
$y = 2t$
Now, differentiation with respect to. x
$\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx} = 2.\frac{-1}{\sqrt{1-x^2}} = \frac{-2}{\sqrt{1-x^2}}$
Therefore, the answer is $\frac{-2}{\sqrt{1-x^2}}$
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NCERT Continuity and Differentiability Class 12 Solutions: Exercise: 5.4
Page number: 130
Total questions: 10
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Question 1: Differentiate the following w.r.t. x:
$\frac{e ^x }{\sin x }$
Answer:
The given function is
$f(x)=\frac{e ^x }{\sin x }$
We differentiate with the help of the Quotient rule.
$f^{'}(x)=\frac{\frac{d(e^x)}{dx}.\sin x-e^x.\frac{(\sin x)}{dx} }{\sin^2 x }$
$=\frac{e^x.\sin x-e^x.\cos }{\sin^2 x } = \frac{e^x(\sin x-\cos x)}{\sin^2x}$
Therefore, the answer is $\frac{e^x(\sin x-\cos x)}{\sin^2x}$
Question 2: Differentiate the following w.r.t. x:
$e ^{\sin ^{-1}x}$
Answer:
The given function is
$f(x)=e ^{\sin ^{-1}x}$
Let $g(x)={\sin ^{-1}x}$
Then,
$f(x)=e^{g(x)}$
Now, differentiation with respect to. x
$f^{'}(x)=g^{'}(x).e^{g(x)}$ -(i)
$g(x) = \sin^{-1}x \rightarrow g^{'}(x ) = \frac{1}{\sqrt{1-x^2}}$
Put this value in our equation (i)
$f^{'}(x) = \frac{1}{\sqrt{1-x^2}}.e^{\sin^{-1}x} = \frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}$
Question 3: Differentiate the following w.r.t. x:
$e ^{x^3}$
Answer:
The given function is
$f(x)=e ^{x^3}$
Let $g(x)=x^3$
Then,
$f(x)=e^{g(x)}$
Now, differentiation with respect to. x
$f^{'}(x)=g^{'}(x).e^{g(x)}$ -(i)
$g(x) = x^3 \rightarrow g^{'}(x ) =3x^2$
Put this value in our equation (i)
$f^{'}(x) =3x^2.e^{x^3}$
Therefore, the answer is $3x^2.e^{x^3}$
Question 4: Differentiate the following w.r.t. x:
$\sin ( \tan ^ { -1} e ^{-x })$
Answer:
The given function is
$f(x)=\sin ( \tan ^ { -1} e ^{-x })$
Let's take $g(x ) = \tan^{-1}e^{-x}$
Now, our function reduces to
$f(x) = \sin(g(x))$
Now,
$f^{'}(x) = g^{'}(x)\cos(g(x))$ -(i)
And
$g(x)=\tan^{-1}e^{-x}\\\rightarrow g^{'}(x) = \frac{d(\tan^{-1}e^{-x})}{dx}.\frac{d(e^{-x})}{dx}= \frac{1}{1+(e^{-x})^2}.-e^{-x} = \frac{-e^{-x}}{1+e^{-2x}}$
Put this value in our equation (i)
$f^{'}(x) =\frac{-e^{-x}}{1+e^{-2x}}\cos(\tan^{-1}e^{-x})$
Therefore, the answer is $\frac{-e^{-x}}{1+e^{-2x}}\cos(\tan^{-1}e^{-x})$
Question 5: Differentiate the following w.r.t. x:
$\log (\cos e ^x )$
Answer:
The given function is
$f(x)=\log (\cos e ^x )$
Let's take $g(x ) = \cos e^{x}$
Now, our function reduces to
$f(x) = \log(g(x))$
Now,
$f^{'}(x) = g^{'}(x).\frac{1}{g(x)}$ -(i)
And
$g(x)=\cos e^{x}\\\rightarrow g^{'}(x) = \frac{d(\cos e^{x})}{dx}.\frac{d(e^{x})}{dx}= (-\sin e^x).e^{x} = -e^x.\sin e^x$
Put this value in our equation (i)
$f^{'}(x) =-e^x.\sin e^x.\frac{1}{\cos e^x} = -e^x.\tan e^x \ \ \ \ \ (\because \frac{\sin x}{\cos x}=\tan x)$
Therefore, the answer is $-e^x.\tan e^x,\ \ \ e^x\neq (2n+1)\frac{\pi}{2},\ \ n\in N$
Question 6: Differentiate the following w.r.t. x:
$e ^x + e ^{x^2} + .....e ^{x^5}$
Answer:
The given function is
$f(x)= e ^x + e ^{x^2} + .....e ^{x^5}$
Now, differentiation w.r.t. x is
$f^{'}(x)= \frac{d(e^x)}{dx}.\frac{d(x)}{dx}+\frac{d(e^{x^2})}{dx}.\frac{d(x^2)}{dx}+\frac{d(e^{x^3})}{dx}.\frac{d(x^3)}{dx}+\frac{d(e^{x^4})}{dx}.\frac{d(x^4)}{dx}+\frac{d(e^{x^5})}{dx}.\frac{d(x^5)}{dx}$
$=e^x.1+e^{x^2}.2x+e^{x^3}.3x^2+e^{x^4}.4x^3+e^{x^5}.5x^4$
$=e^x+2xe^{x^2}+3x^2e^{x^3}+4x^3e^{x^4}+5x^4e^{x^5}$
Therefore, answer is $e^x+2xe^{x^2}+3x^2e^{x^3}+4x^3e^{x^4}+5x^4e^{x^5}$
Question 7: Differentiate the following w.r.t. x:
$\sqrt { e ^{ \sqrt x }} , x > 0$
Answer:
The given function is
$f(x)=\sqrt { e ^{ \sqrt x }}$
Let's take $g(x ) = \sqrt x$
Now, our function reduces to
$f(x) = \sqrt {e^{g(x)}}$
Now,
$f^{'}(x) = g^{'}(x).\frac{1}{2\sqrt{e^{g(x)}}}.\frac{d({e^{g(x)}})}{dx} = g{'}(x).\frac{1}{2\sqrt{e^{g(x)}}}.{e^{g(x)}} = \frac{g^{'}(x).e^{g(x)}}{2.\sqrt{e^{g(x)}}} = \frac{g^{'}(x).e^{\sqrt x}}{2.\sqrt{e^{\sqrt x}}}$ -(i)
And
$g(x)=\sqrt x\\\rightarrow g^{'}(x) = \frac{(\sqrt x)}{dx}=\frac{1}{2\sqrt x}$
Put this value in our equation (i)
$f^{'}(x) =\frac{e^{\sqrt x}}{2\sqrt x.2.\sqrt{e^{\sqrt x}}} = \frac{e^{\sqrt x}}{4\sqrt{xe^{\sqrt x}}}$
Therefore, the answer is $\frac{e^{\sqrt x}}{4\sqrt{xe^{\sqrt x}}}.\ \ x>0$
Question 8: Differentiate the following w.r.t. x: $\log ( \log x ) , x > 1$
Answer:
The given function is
$f(x)=\log ( \log x )$
Let's take $g(x ) = \log x$
Now, our function reduces to
$f(x) = \log(g(x))$
Now,
$f^{'}(x) = g^{'}(x).\frac{1}{g(x)}$ -(i)
And
$g(x)=\log x\\\rightarrow g^{'}(x) = \frac{1}{x}$
Put this value in our equation (i)
$f^{'}(x) =\frac{1}{x}.\frac{1}{\log x} = \frac{1}{x\log x}$
Therefore, the answer is $\frac{1}{x\log x}, \ \ x>1$
Question 9: Differentiate the following w.r.t. x:
$\frac{\cos x }{\log x} , x > 0$
Answer:
The given function is
$f(x)=\frac{\cos x }{\log x}$
We differentiate with the help of the Quotient rule.
$f^{'}(x)=\frac{\frac{d(\cos x)}{dx}.\log x-\cos x.\frac{(\log x)}{dx} }{(\log x)^2 }$
$=\frac{(-\sin x).\log x-\cos x.\frac{1}{x} }{(\log x)^2 } = \frac{-(x\sin x\log x+\cos x)}{x(\log x)^2}$
Therefore, the answer is $\frac{-(x\sin x\log x+\cos x)}{x(\log x)^2}$
Question 10: Differentiate the following w.r.t. x:
$\cos ( log x + e ^x ) , x > 0$
Answer:
The given function is
$f(x)=\cos ( log x + e ^x )$
Let's take $g(x) = ( log x + e ^x )$
Then, our function reduces to
$f(x) = \cos (g(x))$
Now, differentiation w.r.t. x is
$f^{'}(x) = g^{'}(x)(-\sin) (g(x))$ -(i)
And
$g(x) = ( log x + e ^x )$
$g^{'}(x)= \frac{d(\log x)}{dx}+\frac{d(e^x)}{dx} = \frac{1}{x}+e^x$
Put this value in our equation (i)
$f^{'}(x) = -\left ( \frac{1}{x}+e^x \right )\sin (\log x+e^x)$
Therefore, the answer is $-\left ( \frac{1}{x}+e^x \right )\sin (\log x+e^x), x>0$
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NCERT Continuity and Differentiability Class 12 Solutions: Exercise: 5.5
Page number: 134
Total questions: 18
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Question 1: Differentiate the functions w.r.t. x. $\cos x . \cos 2x .\cos 3x$
Answer:
The given function is
$y=\cos x. \cos 2x .\cos 3x$
Now, take a look at both sides.
$\log y=\log (\cos x . \cos 2x .\cos 3x)$
$\log y = \log \cos x + \log \cos 2x + \log \cos 3x$
Now, differentiation with respect to. x
$\log y=\log (\cos x . \cos 2x .\cos 3x)$
$\frac{d(\log y )}{dx} = \frac{\log \cos x}{dx} + \frac{\log \cos 2x}{dx} + \frac{\log \cos 3x}{dx}$
$\frac{1}{y}.\frac{dy}{dx} =$
$(-\sin x)\frac{1}{\cos x}+(-2\sin 2x)\frac{1}{\cos 2x}+(-3\sin3x).\frac{1}{\cos3x}$
$\frac{1}{y} \frac{dy}{dx}= (\tan x+ \tan 2x+ \tan 3x )$
$(\because \frac{\sin x }{\cos x} =\tan x)$
$ \frac{dy}{dx}=-y(\tan x+\tan 2x+\tan 3x)$
$ \frac{dy}{dx}= -\cos x \cos 2x\cos 3x(\tan x+\tan 2x+\tan 3x)$
There, the answer is $-\cos x\cos 2x\cos 3x(\tan x+\tan 2x+\tan 3x)$
Question 2: Differentiate the functions with respect to. x.
$\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}$
Answer:
The given function is
$y=\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}$
Take logs on both sides.
$\log y=\frac{1}{2}\log\left ( \frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)} \right )$
$\log y =$
$\frac{1}{2} (\log(x-1)+\log(x-2)-\log(x-3)-\log(x-4)$
$-\log(x-5))$
Now, differentiation w.r.t. x is
$\frac{d(\log y)}{dx} = \frac{1}{2} (\frac{d(\log(x-1))}{dx}+\frac{d(\log(x-2))}{dx}-\frac{d(\log(x-3))}{dx}$
$-\frac{d(\log(x-4))}{dx}- \frac{d(\log(x-5))}{dx})$
$\frac{1}{y}\frac{dy}{dx}=\frac{1}{2}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})\\ \frac{dy}{dx}$
$=y\frac{1}{2}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})$
$ \frac{dy}{dx} = \frac{1}{2}\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}$
$(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})$
Therefore, the answer is $\frac{1}{2}\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})$
Question 3: Differentiate the functions w.r.t. x. $(\log x ) ^{\cos x}$
Answer:
The given function is
$y=(\log x ) ^{\cos x}$
Take logs on both sides.
$\log y=\cos x\log (\log x )$
Now, differentiation w.r.t x is
$\frac{d(\log y)}{dx}= \frac{d(\cos x\log(\log x))}{dx}$
$\frac{1}{y} \frac{dy}{dx}= (-\sin x)(\log(\log x))+\cos x (\frac{1}{\log x} \cdot \frac{1}{x})$
$\frac{dy}{dx}= y( \cos x.\frac{1}{\log x}.\frac{1}{x}-\sin x\log(\log x) )\\$
$\frac{dy}{dx} = (\log x)^{\cos x}( \frac{\cos x}{x\log x}-\sin x\log(\log x) )$
Therefore, the answer is $(\log x)^{\cos x}( \frac{\cos x}{x\log x}-\sin x\log(\log x) )$
Question 4: Differentiate the functions w.r.t. x. $x ^x - 2 ^{ \sin x }$
Answer:
The given function is
$y = x ^x - 2 ^{ \sin x }$
Let's take $t = x^x$
Take logs on both sides.
$\log t=x\log x\\$
Now, differentiation w.r.t x is
$\log t=x\log x\\ \frac{d(\log t)}{dt}.\frac{dt}{dx} = \frac{d(x\log x)}{dx} \ \ \ \ \ \ \ (by \ chain \ rule)$
$\frac{1}{t}.\frac{dt}{dx} = \log x +1$
$\frac{dt}{dx} = t(\log x+1)$
$\frac{dt}{dx}= x^x(\log x+1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because t = x^x )$
Similarly, take $k = 2^{\sin x}$
Now, take the log on both sides and differentiate with respect to. x
$\log k=\sin x\log 2\\ \frac{d(\log k)}{dk}.\frac{dk}{dx} = \frac{d(\sin x\log 2)}{dx} \ \ \ \ \ \ \ (by \ chain \ rule)$
$\frac{1}{k}.\frac{dk}{dx} = \cos x \log 2$
$\frac{dk}{dx} = k(\cos x \log 2)$
$\frac{dk}{dx}= 2^{\sin x}(\cos x\log 2)$
$(\because k = 2^{\sin x} )$
Now,
$\frac{dy}{dx} = \frac{dt}{dx}-\frac{dk}{dx}$
$\frac{dy}{dx} = x^x(\log x+1 )- 2^{\sin x}(\cos x\log 2)$
Therefore, the answer is $x^x(\log x+1 )- 2^{\sin x}(\cos x \log 2)$
Question 5: Differentiate the functions w.r.t. x. $( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4$
Answer:
The given function is
$y=( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4$
Take logs on both sides.
$\log y=\log [( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4]$
$ \log y = 2\log(x+3)+3\log(x+4)+4\log(x+5)$
Now, differentiate w.r.t. x we get,
$\frac{1}{y}.\frac{dy}{dx} = 2.\frac{1}{x+3}+3.\frac{1}{x+4}+4.\frac{1}{x+5}$
$\frac{dy}{dx}=y\left ( \frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5} \right ) $
$\frac{dy}{dx} = (x+3)^2.(x+4)^3.(x+5)^4.\left ( \frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5} \right )$
$\frac{dy}{dx} = (x+3)^2.(x+4)^3.(x+5)^4.$
$\left ( \frac{2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}{(x+3)(x+4)(x+5)} \right )$
$ \frac{dy}{dx} = (x + 3) (x + 4)^2 (x + 5)^3$
$(9x^2 + 70x + 133)$
Therefore, the answer is $(x + 3) (x + 4)^2 (x + 5)^3 (9x^2 + 70x + 133)$
Question 6: Differentiate the functions w.r.t. x. $( x+ \frac{1}{x} ) ^ x + x ^{ 1 + \frac{1}{x} }$
Answer:
The given function is
$y = ( x+ \frac{1}{x} ) ^ x + x ^{ 1 + \frac{1}{x} }$
Let's take $t = ( x+ \frac{1}{x} ) ^ x$
Now, take a look at both sides.
$\log t =x \log ( x+ \frac{1}{x} )$
Now, differentiate with respect. x
We get,
$\frac{1}{t}.\frac{dt}{dx}=\log \left ( x+\frac{1}{x} \right )+x(1-\frac{1}{x^2}).\frac{1}{\left ( x+\frac{1}{x} \right )}$
$= \frac{x^2-1}{x^2+1}+\log \left ( x+\frac{1}{x} \right )$
$ \frac{dt}{dx} = t(\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))$
$ \frac{dt}{dx} = \left ( x+\frac{1}{x} \right )^x (\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))$
Similarly, take $k = x^{1+\frac{1}{x}}$
Now, take a look at both sides.
$\log k = ({1+\frac{1}{x}})\log x$
Now, differentiate with respect. x
We get,
$\frac{1}{k}.\frac{dk}{dx}=\frac{1}{x} \left ( 1+\frac{1}{x} \right )+(-\frac{1}{x^2}).\log x$
$= \frac{x^2+1}{x^2}+\frac{-1}{x^2}.\log x$
$ \frac{dk}{dx} = t(\frac{x^2+1}{x^2}+\left (\frac{-1}{x^2} \right )\log x)$
$\frac{dk}{dx} = x^{x+\frac{1}{x}}\left (\frac{x^2+1-\log x}{x^2} \right )$
Now,
$\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx} = \left ( x+\frac{1}{x} \right )^x (\left (\frac{x^2-1}{x^2+1} \right )+$
$\log \left ( x+\frac{1}{x} \right ))+x^{x+\frac{1}{x}}\left (\frac{x^2+1-\log x}{x^2} \right )$
Therefore, the answer is $\left ( x+\frac{1}{x} \right )^x (\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))+x^{x+\frac{1}{x}}\left (\frac{x^2+1-\log x}{x^2} \right )$
Question 7: Differentiate the functions with respect to. x. $(\log x )^x + x ^{\log x }$
Answer:
The given function is
$y = (\log x )^x + x ^{\log x }$
Let's take $t = (\log x)^x$
Now, take a look at both sides.
$\log t = x \log(\log x)$
Now, differentiate with respect. x
We get,
$\frac{1}{t}\frac{dt}{dx} = \log (\log x) + x.\frac{1}{x}.\frac{1}{\log x}$
$= \log (\log x)+\frac{1}{\log x}\\$
$ \frac{dt}{dx}= t.(\log (\log x)+\frac{1}{\log x})\\$
$ \frac{dt}{dx} =(\log x)^x(\log (\log x)) + (\log x)^x.\frac{1}{\log x}$
$=(\log x)^x(\log (\log x))+ (\log x )^{x-1}$
Similarly, take $k = x^{\log x}$
Now, take a look at both sides.
$\log k = \log x \log x = (\log x)^2$
Now, differentiate with respect. x
We get,
$\frac{1}{k}\frac{dk}{dx} =2 (\log x).\frac{1}{x} \\$
$ \frac{dt}{dx}= k.\left ( 2 (\log x).\frac{1}{x} \right )\\$
$ \frac{dt}{dx} = x^{\log x}.\left (2 (\log x).\frac{1}{x} \right ) = 2x^{\log x-1}.\log x$
Now,
$\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx} =(\log x)^x(\log (\log x))+ (\log x )^{x-1}+ 2x^{\log x-1}.\log x$
Therefore, the answer is $(\log x)^x(\log (\log x))+ (\log x )^{x-1}+ 2x^{\log x-1}.\log x$
Question 8: Differentiate the functions with respect to. x. $(\sin x )^x + \sin ^{-1} \sqrt x$
Answer:
The given function is
$(\sin x )^x + \sin ^{-1} \sqrt x$
Let's take $t = (\sin x)^x$
Now, take a look at both sides.
$\log t = x \log(\sin x)$
Now, differentiate with respect. x
We get,
$\frac{1}{t}\frac{dt}{dx} = \log (\sin x) + x.\cos x.\frac{1}{\sin x}$
$= \log (\sin x)+x.\cot x$
$ \ \ \ (\because \frac{\cos x}{\sin x}=\cot x)\\$
$ \frac{dt}{dx}= t.(\log (\sin x)+x.\cot x)\\$
$ \frac{dt}{dx} =(\sin x)^x(\log (\sin x)+x\cot x)$
Similarly, take $k = \sin^{-1}\sqrt x$
Now, differentiate with respect. x
We get,
$\frac{dk}{dt} = \frac{1}{\sqrt{1-(\sqrt x)^2}}.\frac{1}{2\sqrt x}$
$= \frac{1}{2\sqrt{x-x^2}}\\$
$ \frac{dk}{dt}=\frac{1}{2\sqrt{x-x^2}}\\$
Now,
$\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx} =(\sin x)^x(\log (\sin x)+x\cot x)+\frac{1}{2\sqrt{x-x^2}}$
Therefore, the answer is $(\sin x)^x(\log (\sin x)+x\cot x)+\frac{1}{2\sqrt{x-x^2}}$
Question 9: Differentiate the functions w.r.t. x $y=x^{\sin x}+(\sin x)^{\cos x}$
Answer:
The given function is
$y=x^{\sin x}+(\sin x)^{\cos x}$
Now, take $t = x^{\sin x}$
Now, take a look at both sides.
$\log t = \sin x \log x$
Now, differentiate it w.r.t. x
We get,
$\frac{1}{t}\frac{dt}{dx} = \cos x \log x+\frac{1}{x}.\sin x\\$
$\frac{dt}{dx}=t\left ( \cos x \log x+\frac{1}{x}.\sin x \right )\\$
$\frac{dt}{dx}= x^{\sin x}\left ( \cos x \log x+\frac{1}{x}.\sin x \right )$
Similarly, take $k = (\sin x)^{\cos x}$
Now, take a look at both sides.
$\log k = \cos x \log (\sin x)$
Now, differentiate it w.r.t. x
We get,
$\frac{1}{k}\frac{dk}{dt} = (-\sin x)(\log (\sin x)) + \cos x.\frac{1}{\sin x}.\cos x$
$=-\sin x\log(\sin x)+\cot x.\cos x\\ \frac{dk}{dt}$
$= k\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )\\ $
$\frac{dk}{dt}=(\sin x)^{\cos x}\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )$
Now,
$\frac{dy}{dx} = x^{\sin x}(\cos x \log x+\frac{1}{x}.\sin x)+$
$(\sin x)^{\cos x} ( -\sin x\log(\sin x)+\cot x.\cos x)$
Therefore, the answer is $x^{\sin x}\left (\cos x\log x+\frac{1}{x}.\sin x \right )+$
$(\sin x)^{\cos x}\left (-\sin x\log(\sin x)+\cot x.\cos x\right )$
Question 10: Differentiate the functions with respect to. x. $x ^ {x \cos x} + \frac{x^2 + 1 }{x^2 -1 }$
Answer:
The given function is
$y=x^{\sin x}+(\sin x)^{\cos x}$
Now, take $t = x^{\sin x}$
Now, take a look at both sides.
$\log t = \sin x \log x$
Now, differentiate it w.r.t. x
We get,
$\frac{1}{t}\frac{dt}{dx} = \cos x \log x+\frac{1}{x}.\sin x\\$
$\frac{dt}{dx}=t\left ( \cos x \log x+\frac{1}{x}.\sin x \right )\\$
$\frac{dt}{dx}= x^{\sin x}\left ( \cos x \log x+\frac{1}{x}.\sin x \right )$
Similarly, take $k = (\sin x)^{\cos x}$
Now, take a look at both sides.
$\log k = \cos x \log (\sin x)$
Now, differentiate it w.r.t. x
We get,
$\frac{1}{k}\frac{dk}{dt} = (-\sin x)(\log (\sin x)) + \cos x.\frac{1}{\sin x}.\cos x$
$=-\sin x\log(\sin x)+\cot x.\cos x\\ \frac{dk}{dt}$
$= k\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )\\ $
$\frac{dk}{dt}=(\sin x)^{\cos x}\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )$
Now,
$\frac{dy}{dx} = x^{\sin x}(\cos x \log x+\frac{1}{x}.\sin x)+$
$(\sin x)^{\cos x} ( -\sin x\log(\sin x)+\cot x.\cos x)$
Therefore, the answer is $x^{\sin x}\left (\cos x\log x+\frac{1}{x}.\sin x \right )+$
$(\sin x)^{\cos x}\left (-\sin x\log(\sin x)+\cot x.\cos x\right )$
Question 11: Differentiate the functions w.r.t. x. $( x \cos x )^ x + ( x \sin x )^{1/ x}$
Answer:
Given function is
$f(x)=( x \cos x)^ x + ( x \sin x )^{1/ x}$
Let's take $t = (x\cos x)^x$
Now, take a look at both sides.
$\log t =x\log (x\cos x) = x(\log x+\log \cos x)$
Now, differentiate w.r.t. x
we get,
$\frac{1}{t}\frac{dt}{dx} =(\log x+\log \cos x)+x(\frac{1}{x}+\frac{1}{\cos x}.(-\sin x))\\$
$\frac{dt}{dx} = t(\log x + \log \cos x+1-x\tan x) $
$\ \ \ \ \ \ \ \ \ (\because \frac{\sin x}{\cos x}= \tan x)\\$
$\frac{dt}{dx}= (x\cos x)^x(\log x + \log \cos x+1-x\tan x)\\$
$\frac{dt}{dx}=(x\cos x)^x(+1-x\tan x+\log (x\cos x))$
Similarly, take $k = (x\sin x)^{\frac{1}{x}}$
Now, take a look at both sides.
$\log k = \frac{1}{x}(\log x+\log \sin x)$
Now, differentiate w.r.t. x
we get,
$\frac{1}{k}\frac{dk}{dx} =(\frac{-1}{x^2})(\log x+\log \sin x)+$
$\frac{1}{x}(\frac{1}{x}+\frac{1}{\sin x}.(\cos x))\\$
$\frac{dk}{dx} = \frac{k}{x^2}(-\log x - \log \sin x+\frac{1}{x^2}+\frac{\cot x}{x})$
$\ \ \ \ \ \ \ \ \ (\because \frac{\cos x}{\sin x}= \cot x)\\$
$\frac{dk}{dx}=\frac{(x\sin x)^{\frac{1}{x}}}{x^2}(-\log x - \log \sin x+\frac{1}{x^2}+\frac{\cot x}{x})\\$
$\frac{dk}{dx}=(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}$
Now,
$\frac{dy}{dx}= \frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx}= (x\cos x)^x(+1-x\tan x+\log (x\cos x))+$
$(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}$
Therefore, the answer is $(x\cos x)^x(1-x\tan x+\log (x\cos x))+$
$(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}$
Question 12: Find $\frac{dy}{dx}$ of the functions given in Exercises 12 to 15
$x ^ y + y ^ x = 1$ .
Answer:
The given function is
$f(x)=x ^ y + y ^ x = 1$
Now, take $t = x^y$
Take logs on both sides.
$\log t = y\log x$
Now, differentiate w.r.t x
We get,
$\frac{1}{t}\frac{dt}{dx} = \frac{dy}{dx}(\log x)+y\frac{1}{x}=\frac{dy}{dx}(\log x)+\frac{y}{x}\\$
$ \frac{dt}{dx}= t(\frac{dy}{dx}(\log x)+\frac{y}{x})\\$
$ \frac{dt}{dx}= ( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x})$
Similarly, take $k = y^x$
Now, take a look at both sides.
$\log k = x\log y$
Now, differentiate with respect. x
We get,
$\frac{1}{k}\frac{dk}{dx} = (\log y)+x\frac{1}{y}\frac{dy}{dx}=\log y+\frac{x}{y}\frac{dy}{dx}\\$
$ \frac{dk}{dx}= k(\log y+\frac{x}{y}\frac{dy}{dx})\\$
$ \frac{dk}{dx}= (y^x)(\log y+\frac{x}{y}\frac{dy}{dx})$
Now,
$f^{'}(x)= \frac{dt}{dx}+\frac{dk}{dx}= 0$
$( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x}) + (y^x)(\log y+\frac{x}{y}\frac{dy}{dx}) = 0\\$
$\frac{dy}{dx}(x^y(\log x)+xy^{x-1}) = -(yx^{y-1}+y^x(\log y))\\ $
$\frac{dy}{dx}= \frac{ -(yx^{y-1}+y^x(\log y))}{(x^y(\log x)+xy^{x-1})}$
Therefore, the answer is $\frac{ -(yx^{y-1}+y^x(\log y))}{(x^y(\log x)+xy^{x-1})}$
Question 13: Find $\frac{dy}{dx}$ of the functions given in Exercises 12 to 15.
$y^x = x ^y$
Answer:
The given function is
$f(x)\rightarrow x ^ y = y ^ x$
Now, take $t = x^y$
Take logs on both sides.
$\log t = y\log x$
Now, differentiate w.r.t x
We get,
$\frac{1}{t}\frac{dt}{dx} = \frac{dy}{dx}(\log x)+y\frac{1}{x}=\frac{dy}{dx}(\log x)+\frac{y}{x}\\$
$ \frac{dt}{dx}= t(\frac{dy}{dx}(\log x)+\frac{y}{x})$
$ \frac{dt}{dx}= ( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x})$
Similarly, take $k = y^x$
Now, take a look at both sides.
$\log k = x\log y$
Now, differentiate with respect. x
We get,
$\frac{1}{k}\frac{dk}{dx} = (\log y)+x\frac{1}{y}\frac{dy}{dx}=\log y+\frac{x}{y}\frac{dy}{dx}\\$
$ \frac{dk}{dx}= k(\log y+\frac{x}{y}\frac{dy}{dx})\\ $
$\frac{dk}{dx}= (y^x)(\log y+\frac{x}{y}\frac{dy}{dx})$
Now,
$f^{'}(x)\rightarrow \frac{dt}{dx}= \frac{dk}{dx}$
$( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x}) = (y^x)(\log y+\frac{x}{y}\frac{dy}{dx})\\$
$ \frac{dy}{dx}(x^y(\log x)-xy^{x-1}) = (y^x(\log y)-yx^{y-1})\\$
$ \frac{dy}{dx}= \frac{ y^x(\log y)-yx^{y-1}}{(x^y(\log x)-xy^{x-1})} $
$= \frac{x}{y}\left ( \frac{y-x\log y}{x-y\log x}\right )$
Therefore, the answer is $\frac{x}{y}\left ( \frac{y-x\log y}{x-y\log x}\right )$
Question 14: Find $\frac{dy}{dx}$ of the functions given in Exercises 12 to 15. $(\cos x )^y = ( \cos y )^x$
Answer:
The given function is
$f(x)\rightarrow (\cos x) ^ y = (\cos y) ^ x$
Now, take the log on both sides.
$y\log \cos x = x \log \cos y$
Now, differentiate w.r.t x
$\frac{dy}{dx}(\log \cos x)-y\tan x = \log \cos y-x\tan y\frac{dy}{dx}$
By taking similar terms on the same side
We get,
$(\frac{dy}{dx}(\log \cos x)-y\tan x)$
$= (\log \cos y-x\tan y\frac{dy}{dx})\\ \frac{dy}{dx} \left (\log \cos x+(\cos y)^x.x\tan y) \right )$
$= \left ( \log \cos y+(\cos x)^y.y\tan x \right )\\ \frac{dy}{dx}$
$= \frac{\left (\log \cos y+y\tan x \right )}{\left ( \log \cos x+x\tan y) \right )}$
$= \frac{y\tan x+\log \cos y}{x\tan y+\log \cos x}$
Therefore, the answer is $\frac{y\tan x+\log \cos y}{x\tan y+\log \cos x}$
Question 15: Find $\frac{dy}{dx}$ of the functions given in Exercises 12 to 15. $xy = e ^{x-y}$
Answer:
The given function is
$f(x)\rightarrow xy = e ^{x-y}$
Now, take a look at both sides.
$\log x+\ log y = (x-y)(1) \ \ \ \ \ \ \ \ \ \ \ \ (\because \log e = 1)\\ \log x+\ log y = (x-y)$
Now, differentiate w.r.t x
$\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}$
By taking similar terms on the same side
We get,
$(\frac{1}{y}+1)\frac{dy}{dx}=1-\frac{1}{x}\\ \frac{y+1}{y}.\frac{dy}{dx}= \frac{x-1}{x}\\ \frac{dy}{dx}= \frac{y}{x}.\frac{x-1}{y+1}$
Therefore, the answer is $\frac{y}{x}.\frac{x-1}{y+1}$
Question 16: Find the derivative of the function given by $f (x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$ and hence find
f ' (1)
Answer:
The given function is
$y = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$
Take logs on both sides.
$\log y =\log (1 + x) + \log (1 + x^2) +\log (1 + x^4) +\log (1 + x^8)$
NOW, differentiate with respect. x
$\frac{1}{y}.\frac{dy}{dx} = \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8}\\$
$ \frac{dy}{dx}=y.\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )\\ $
$\frac{dy}{dx}= (1 + x) (1 + x^2) (1 + x^4) (1 + x^8).\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )$
Therefore, $f^{'}(x)=$
$ (1 + x) (1 + x^2) (1 + x^4) (1 + x^8).\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )$
Now, the value of $f^{'}(1)$ is
$f^{'}(1)= (1 + 1) (1 + 1^2) (1 + 1^4) (1 + 1^8).\left ( \frac{1}{1+1}+ \frac{2(1)}{1+1^2}+\frac{4(1)^3}{1+1^4}+ \frac{8(1)^7}{1+1^8} \right )\\$
$ f^{'}(1)=16.\frac{15}{2} = 120$
Question 17(1): Differentiate $(x^2 - 5x + 8) (x^3 + 7x + 9)$ in three ways mentioned below:
(i) By using the product rule
Answer:
The given function is
$f(x)=(x^2 - 5x + 8) (x^3 + 7x + 9)$
Now, we need to differentiate using the product rule.
$f^{'}(x)=\frac{d((x^2 - 5x + 8))}{dx}. (x^3 + 7x + 9)+(x^2 - 5x + 8).\frac{d( (x^3 + 7x + 9))}{dx}\\$
$= (2x-5).(x^3+7x+9)+(x^2-5x+8)(3x^2+7)\\$
$ =2x^4+14x^2+18x-5x^3-35x-45+3x^4-15x^3+24x^2+7x^2-35x+56\\$
$ = 5x^4 -20x^3+45x^2-52x+11$
Therefore, the answer is $5x^4 -20x^3+45x^2-52x+11$
Question 17(2): Differentiate $(x^2 - 5x + 8) (x^3 + 7x + 9)$ in three ways mentioned below:
(ii) by expanding the product to obtain a single polynomial.
Answer:
The given function is
$f(x)=(x^2 - 5x + 8) (x^3 + 7x + 9)$
Multiply both to obtain a single higher-degree polynomial.
$f(x) = x^2(x^3+7x+9)-5x(x^3+7x+9)+8(x^3+7x+9)$
$= x^5+7x^3+9x^2-5x^4-35x^2-45x+8x^3+56x+72$
$= x^5-5x^4+15x^3-26x^2+11x+72$
Now, differentiate with respect. x
We get,
$f^{'}(x)=5x^4-20x^3+45x^2-52x+11$
Therefore, the answer is $5x^4-20x^3+45x^2-52x+11$
Question 17(3): Differentiate $(x^2 - 5x + 8) (x^3 + 7x + 9)$ in three ways mentioned below:
(iii) by logarithmic differentiation.
Do they all give the same answer?
Answer:
The given function is
$y=(x^2 - 5x + 8) (x^3 + 7x + 9)$
Now, take a look at both sides.
$\log y = \log (x^2-5x+8)+\log (x^3+7x+9)$
Now, differentiate with respect. x
We get,
$\frac{1}{y}.\frac{dy}{dx} = \frac{1}{x^2-5x+8}.(2x-5) + \frac{1}{x^3+7x+9}.(3x^2+7)\\$
$ \frac{dy}{dx}= y.\left ( \frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)} \right )\\$
$ \frac{dy}{dx}=(x^2-5x+8)(x^3+7x+9).\left ( \frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)} \right )\\$
$ \frac{dy}{dx} = (2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)\\ \frac{dy}{dx} = 5x^4-20x^3+45x^2-56x+11$
Therefore, the answer is $5x^4-20x^3+45x^2-56x+11$
And yes, they all give the same answer.
Question 18: If u, v and w are functions of x, then show that $\frac{d}{dx} ( u,v,w) = \frac{du}{dx} v.. w +u. \frac{dv }{dx } v. w+ u . \frac{dv}{dx } . w+u.v \frac{dw}{dx}$ in two ways - first by repeated application of the product rule, second by logarithmic differentiation.
Answer:
It is given that u, v and w are the functions of x
Let $y = u.v.w$
Now, we differentiate using the product rule with respect to x
First, take $y = u.(vw)$
Now,
$\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{d(v.w)}{dx}.u$ -(i)
Now, again, by the product rule
$\frac{d(v.w)}{dx}= \frac{dv}{dx}.w + \frac{dw}{dx}.v$
Put this in equation (i)
We get,
$\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{dv}{dx}.(u.w) + \frac{dw}{dx}.(u.v)$
Hence, by the product rule, we proved it.
Now, by taking the log
Again take $y = u.v.w$
Now, take a look at both sides.
$\log y = \log u + \log v + \log w$
Now, differentiate with respect. x
We get,
$\frac{1}{y}.\frac{dy}{dx} = \frac{1}{u}.\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}.\frac{dw}{dx}\\ \frac{dy}{dx}= y. \left ( \frac{v.w.\frac{du}{dx}+u.w.\frac{dv}{dx}+u.v.\frac{dw}{dx}}{u.v.w} \right )\\ \frac{dy}{dx} $
$= (u.v.w)\left ( \frac{v.w.\frac{du}{dx}+u.w.\frac{dv}{dx}+u.v.\frac{dw}{dx}}{u.v.w} \right )\\$
$\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{dv}{dx}.(u.w) + \frac{dw}{dx}.(u.v)$
Hence, we proved it by taking the log.
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NCERT Continuity and Differentiability Class 12 Solutions: Exercise: 5.6
Page number: 137
Total questions: 11
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Question 1: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\frac{dy}{dx}$.
$x = 2at^2, y = at^4$
Answer:
The given equations are
$x = 2at^2, y = at^4$
Now, differentiate both with respect to t
We get,
$\frac{dx}{dt}=\frac{d(2at^2)}{dt}= 4at$
Similarly,
$\frac{dy}{dt}=\frac{d(at^4)}{dt}= 4at^3$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{4at^3}{4at} = t^2$
Therefore, the answer is $\frac{dy}{dx}= t^2$
Question 2: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\frac{dy}{dx}$.
Answer:
The given equations are
$x= a \cos \theta , y = b \cos \theta$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(a\cos \theta)}{d\theta}= -a\sin \theta$
Similarly,
$\frac{dy}{d\theta}=\frac{d(b\cos \theta)}{d\theta}= -b\sin \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{-b\sin \theta}{-a\sin \theta} = \frac{b}{a}$
Therefore, answer is $\frac{dy}{dx}= \frac{b}{a}$
Question 3: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\frac{dy}{dx}$. $x = \sin t , y = \cos 2 t$
Answer:
The given equations are
$x = \sin t , y = \cos 2 t$
Now, differentiate both w.r.t t
We get,
$\frac{dx}{dt}=\frac{d(\sin t)}{dt}= \cos t$
Similarly,
$\frac{dy}{dt}=\frac{d(\cos 2t)}{dt}= -2\sin 2t = -4\sin t \cos t $
$\ \ \ \ \ (\because \sin 2x = \sin x\cos x)$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{-4\sin t \cos t }{\cos t} = -4\sin t$
Therefore, the answer is $\frac{dy}{dx} = -4\sin t$
Question 4: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\ \frac {dy}{dx}$
$x = 4t , y = 4/t$
Answer:
The given equations are
$x = 4t , y = 4/t$
Now, differentiate both w.r.t t
We get,
$\frac{dx}{dt}=\frac{d(4 t)}{dt}= 4$
Similarly,
$\frac{dy}{dt}=\frac{d(\frac{4}{t})}{dt}= \frac{-4}{t^2}$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{ \frac{-4}{t^2} }{4} = \frac{-1}{t^2}$
Therefore, the answer is $\frac{dy}{dx} = \frac{-1}{t^2}$
Question 5: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\frac{dy}{dx}$ $x = \cos \theta - \cos 2\theta, y = \sin \theta - \sin 2 \theta$
Answer:
The given equations are
$x = \cos \theta - \cos 2\theta , y = \sin \theta - \sin 2 \theta$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(\cos \theta-\cos 2\theta)}{d\theta}= -\sin \theta -(-2\sin 2\theta) = 2\sin 2\theta - \sin \theta$
Similarly,
$\frac{dy}{d\theta}=\frac{d(\sin \theta - \sin 2\theta)}{d\theta}= \cos \theta -2\cos2 \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{\cos \theta -2\cos 2 \theta}{2\sin2\theta-\sin \theta}$
Therefore, answer is $\frac{dy}{dx}= \frac{\cos \theta -2\cos 2 \theta}{2\sin2\theta-\sin \theta}$
Question 6: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find $\frac{dy}{dx}$ $x = a ( \theta - \sin \theta ), y = a ( 1+ \cos \theta )$
Answer:
Given equations are
$x = a ( \theta - \sin \theta ) , y = a ( 1+ \cos \theta )$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(a(\theta- \sin \theta))}{d\theta}= a(1-\cos \theta)$
Similarly,
$\frac{dy}{d\theta}=\frac{d(a(1+\cos \theta))}{d\theta}=-a\sin \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{-a\sin \theta}{a(1-\cos \theta)} = \frac{-\sin }{1-\cos \theta} =- \cot \frac{\theta}{2} \ \ \ \ \ \ \ (\cot \frac{x}{2}=\frac{\sin x}{1-\cos x})$
Therefore, the answer is $\frac{dy}{dx}=-\cot \frac{\theta}{2}$
Question 7: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find $\frac{dy}{dx}$ $x = \frac{\sin ^3 t }{\sqrt {\cos 2t }}, y = \frac{\cos ^3 t }{\sqrt {\cos 2t }}$
Answer:
Given equations are
$x = \frac{\sin ^3 t }{\sqrt {\cos 2t }} , y = \frac{\cos ^3 t }{\sqrt {\cos 2t }}$
Now, differentiate both w.r.t
We get,
$\frac{dx}{dt}=\frac{d(\frac{\sin ^3 t }{\sqrt {\cos 2t }})}{dt}$
$=\frac{\sqrt{\cos 2t}.\frac{d(\sin^3t)}{dt}-\sin^3t.\frac{d(\sqrt{\cos2t})}{dt}}{(\sqrt{\cos2t})^2}$
$=\frac{3\sin^2 t\cos t.\sqrt{\cos 2t}-\sin^3t.\frac{1}{2\sqrt{\cos 2t}}.(-2\sin 2t)}{{\cos 2t}}$
$=\frac{3\sin^2t\cos t . \cos 2t+sin^3t\sin 2t}{\cos2t\sqrt{\cos2t}}$
$=\frac{\sin^3t\sin2t(3\cot t \cot 2t+1)}{\cos2t\sqrt{\cos2t}} \ \ \ \ \ (\because \frac{\cos }{\sin x}=\cot x)$
Similarly,
$\frac{dy}{dt}=\frac{d( \frac{\cos ^3 t }{\sqrt {\cos 2t }})}{dt}$
$=\frac{\sqrt{\cos 2t}.\frac{d(\cos^3t)}{dt}-\cos^3t.\frac{d(\sqrt{\cos2t})}{dt}}{(\sqrt{\cos2t})^2}$
$=\frac{3\cos^2 t(-\sin t).\sqrt{\cos 2t}-\cos^3t.\frac{1}{2\sqrt{\cos 2t}}.(-2\sin 2t)}{(\sqrt{\cos 2t})^2}$
$=\frac{-3\cos^2t\sin t\cos2t+\cos^3t\sin 2t}{\cos2t\sqrt{\cos2t}}$
$=\frac{ \sin2t\cos^3t(1-3\tan t \cot 2t)}{\cos2t\sqrt{\cos2t}}$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$= \frac{\frac{ \sin2t\cos^3t(1-3\tan t \cot 2t)}{\cos2t\sqrt{\cos2t}} }{\frac{\sin^3t\sin2t(3\cot t \cot 2t+1)}{\cos2t\sqrt{\cos2t}}}$
$= \frac{\cot^3t(1-3\tan t \cot 2t)}{(3\cot t \cot 2t+1)}$
$= \frac{\cos^3t(1-3.\frac{\sin t}{\cos t}.\frac{\cos2t}{\sin 2t})}{\sin^3t(3.\frac{\cos t}{\sin t}.\frac{\cos 2t}{\sin 2t}+1)} $
$= \frac{\cos^2t(\cos t\sin2t -3\sin t \cos 2t)}{\sin^2t(3\cos t \cos2t+\sin t \sin 2t)}$
$=\frac{\cos^2t(\cos t .2\sin t \cos t - 3\sin t (2\cos^2t-1))}{\sin^2t(3\cos t(1-2\sin^2 2t)+\sin t.2\sin t \cos t)}$
$(\because \sin 2x = 2\sin x\cos x \ and \ \cos 2x = 2\cos^2x-1 \ and \ \cos 2x = 1-2\sin^2x)$
$=\frac{\cos^2t(2\sin t\cos^2 t-6\sin t\cos^2t+3\sin t)}{\sin^2t(3\cos t-6\cos t \sin^2t+2\sin^2\cos t)}\\=\frac{sint cost(-4cos^3t+3cost)}{sintcost(3sint-4sin^3t)}$
$\frac{dy}{dx} = \frac{-4\cos^3t+3\cos t}{3\sin t -4\sin^3 t}= \frac{-\cos 3t}{\sin 3t} = -\cot 3t$ $\left ( \because \sin3t = 3\sin t-4\sin^3t \\ \ and \ \cos3t = 4\cos^3t - 3\cos t \right )$
Therefore, the answer is $\frac{dy}{dx} = -\cot 3t$
Question 8: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\frac{dy}{dx}$ $x = a ( \cos t + \log \tan t/2 ),y = a \sin t$
Answer:
Given equations are
$x = a ( \cos t + \log \tan \frac{t}{2} ),y = a \sin t$
Now, differentiate both w.r.t t
We get,
$\frac{dx}{dt}=\frac{d(a ( \cos t + \log \tan \frac{t}{2} ))}{dt}= a(-\sin t + \frac{1}{\tan\frac{t}{2}}.\sec^2\frac{t}{2}.\frac{1}{2})$
$= a(-\sin t+\frac{1}{2}.\frac{\cos \frac{t}{2}}{\sin\frac{t}{2}}.\frac{1}{\cos^2\frac{t}{2}}) = a(-\sin t+\frac{1}{2\sin \frac{t}{2}\cos \frac{t}{2}})$
$=a(-\sin t+\frac{1}{\sin 2.\frac{t}{2}} ) = a(\frac{-\sin^2t+1}{\sin t})= a(\frac{\cos^2t}{\sin t})$
Similarly,
$\frac{dy}{dt}=\frac{d(a\sin t)}{dt}= a\cos t$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{a \cos t }{ a(\frac{\cos^2t}{\sin t})} = \frac{\sin t}{\cos t} = \tan t$
Therefore, the answer is $\frac{dy}{dx} = \tan t$
Question 9: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\frac{dy}{dx}$ $x = a \sec \theta, y = b \ tan \theta$
Answer:
Given equations are
$x = a \sec \theta , y = b \ tan \theta$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(a\sec \theta)}{d\theta}= a\sec \theta \tan \theta$
Similarly,
$\frac{dy}{d\theta}=\frac{d(b\tan \theta)}{d\theta}= b\sec^2 \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{b\sec^2 \theta}{a\sec\theta\tan \theta} = \frac{b\sec\theta}{a\tan \theta}= \frac{b\frac{1}{\cos\theta}}{a\frac{\sin \theta}{\cos \theta}} = \frac{b }{a\sin \theta} = \frac{b cosec \theta}{a}$
Therefore, the answer is $\frac{dy}{dx} = \frac{b cosec \theta}{a}$
Question 10: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find $\frac{dy}{dx}$ $x = a ( \cos \theta + \theta \sin \theta ), y = a ( \sin \theta - \theta \cos \theta )$
Answer:
The given equations are
$x = a ( \cos \theta + \theta \sin \theta ) , y = a ( \sin \theta - \theta \cos \theta )$
Now, differentiate both with respect to $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(a(\cos \theta+ \theta\sin \theta))}{d\theta}= a(-\sin \theta+\sin \theta+ \theta\cos \theta)= a \theta\cos \theta$
Similarly,
$\frac{dy}{d\theta}=\frac{d(a(\sin \theta- \theta\cos \theta))}{d\theta}= a(\cos \theta-\cos \theta+ \theta\sin \theta) = a \theta\sin \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{a \theta\sin \theta}{a \theta\cos \theta} = \tan \theta$
Therefore, the answer is $\frac{dy}{dx}= \tan \theta$
Question 11: If $x = \sqrt {a ^{\sin ^{-1}t}} , y = \sqrt { a ^{ \cos ^{-1}t}}$ , show that $\frac{dy}{dx}$ = - y /x$
Answer:
The given equations are
$x = \sqrt {a ^{\sin ^{-1}t}} , y = \sqrt { a ^{ \cos ^{-1}t}}$
$xy=\sqrt{a^{sin^{-1}t+cos^{-1}t}}\\since\sin ^{-1}x+cos^{-1}x=\frac{\pi}{2}\\xy=a^{\frac{\pi}{2}}=constant=c$
Differentiating with respect to x
$x\frac{dy}{dx}+y=0\\\frac{dy}{dx}=\frac{-y}{x}$
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NCERT Continuity and Differentiability Class 12 Solutions: Exercise: 5.7
Page number: 139
Total questions: 17
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Question 1: Find the second-order derivatives of the functions given in Exercises 1 to 10.
$x^2 + 3x+ 2$
Answer:
The given function is
$y=x^2 + 3x+ 2$
Now, differentiation with respect to. x
$\frac{dy}{dx}= 2x+3$
Now, the second-order derivative
$\frac{d^2y}{dx^2}= 2$
Therefore, the second order derivative is $\frac{d^2y}{dx^2}= 2$
Question 2: Find the second-order derivatives of the functions given in Exercises 1 to 10.
$x ^{20}$
Answer:
The given function is
$y=x ^{20}$
Now, differentiation with respect to. x
$\frac{dy}{dx}= 20x^{19}$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}= 20.19x^{18}= 380x^{18}$
Therefore, second-order derivative is $\frac{d^2y}{dx^2}= 380x^{18}$
Question 3: Find the second-order derivatives of the functions given in Exercises 1 to 10.
$x \cos x$
Answer:
The given function is
$y = x \cos x$
Now, differentiation with respect to. x
$\frac{dy}{dx}= \cos x + x(-\sin x ) = \cos x-x\sin x$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}= -\sin x-(\sin x+x\cos x) = -2\sin x - x\sin x$
Therefore, the second-order derivative is $\frac{d^2y}{dx^2}= -2\sin x - x\sin x$
Question 4: Find the second-order derivatives of the functions given in Exercises 1 to 10.
$\log x$
Answer:
The given function is
$y=\log x$
Now, differentiation with respect to. x
$\frac{dy}{dx}=\frac{1}{x}$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}= \frac{-1}{x^2}$
Therefore, second order derivative is $\frac{d^2y}{dx^2}= \frac{-1}{x^2}$
Question 5: Find the second-order derivatives of the functions given in Exercises 1 to 10.
$x ^3 \log x$
Answer:
The given function is
$y=x^3\log x$
Now, differentiation with respect. x
$\frac{dy}{dx}=3x^2.\log x+x^3.\frac{1}{x}= 3x^2.\log x+ x^2$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}= 6x.\log x+3x^2.\frac{1}{x}+2x=6x.\log x+3x+2x = x(6.\log x+5)$
Therefore, the second-order derivative is $\frac{d^2y}{dx^2} = x(6.\log x+5)$
Question 6: Find the second-order derivatives of the functions given in Exercises 1 to 10.
$e ^x \sin5 x$
Answer:
Given function is
$y= e^x\sin 5x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=e^x.\sin 5x +e^x.5\cos 5x = e^x(\sin5x+5\cos5x)$
Now, the second order derivative is
$\frac{d^2y}{dx^2}= e^x(\sin5x+5\cos5x)+e^x(5\cos5x+5.(-5\sin5x))$
$= e^x(\sin5x+5\cos5x)+e^x(5\cos5x-25\sin5x)=e^x(10\cos5x-24\sin5x)$
$=2e^x(5\cos5x-12\sin5x)$
Therefore, second order derivative is $\frac{dy}{dx}=2e^x(5\cos5x-12\sin5x)$
Question 7: Find the second-order derivatives of the functions given in Exercises 1 to 10.
$e ^{6x}\cos 3x$
Answer:
Given function is
$y= e^{6x}\cos 3x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=6e^{6x}.\cos 3x +e^{6x}.(-3\sin 3x)= e^{6x}(6\cos 3x-3\sin 3x)$
Now, the second order derivative is
$\frac{d^2y}{dx^2}= 6e^{6x}(6\cos3x-3\sin3x)+e^{6x}(6.(-3\sin3x)-3.3\cos3x)$
$= 6e^{6x}(6\cos3x-3\sin3x)-e^{6x}(18\sin3x+9\cos3x)$
$e^{6x}(27\cos3x-36\sin3x) = 9e^{6x}(3\cos3x-4\sin3x)$
Therefore, second order derivative is $\frac{dy}{dx} = 9e^{6x}(3\cos3x-4\sin3x)$
Question 8: Find the second-order derivatives of the functions given in Exercises 1 to 10.
$\tan ^{-1} x$
Answer:
Given function is
$y = \tan^{-1}x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(\tan^{-1}x)}{dx}=\frac{1}{1+x^2}$
Now, the second order derivative is
$\frac{d^2y}{dx^2}= \frac{-1}{(1+x^2)^2}.2x = \frac{-2x}{(1+x^2)^2}$
Therefore, second order derivative is $\frac{d^2y}{dx^2} = \frac{-2x}{(1+x^2)^2}$
Question 9: Find the second-order derivatives of the functions given in Exercises 1 to 10.
$\log (\log x )$
Answer:
Given function is
$y = \log(\log x)$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(\log(\log x))}{dx}=\frac{1}{\log x}.\frac{1}{x}= \frac{1}{x\log x}$
Now, the second order derivative is
$\frac{d^2y}{dx^2}= \frac{-1}{(x\log x)^2}.(1.\log x+x.\frac{1}{x}) = \frac{-(\log x+1)}{(x\log x)^2}$
Therefore, second order derivative is $\frac{d^2y}{dx^2} = \frac{-(\log x+1)}{(x\log x)^2}$
Question 10: Find the second-order derivatives of the functions given in Exercises 1 to 10.
$\sin (\log x )$
Answer:
Given function is
$y = \sin(\log x)$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(\sin(\log x))}{dx}=\cos (\log x).\frac{1}{x}= \frac{\cos (\log x)}{x}$
Now, the second order derivative is
Using Quotient rule
$\frac{d^2y}{dx^2}=\frac{-\sin(\log x)\frac{1}{x}.x-\cos(\log x).1}{x^2} = \frac{-(\sin (\log x)+\cos(\log x))}{x^2}$
Therefore, second order derivative is $\frac{d^2y}{dx^2} = \frac{-(\sin (\log x)+\cos(\log x))}{x^2}$
Question 11: If $y = 5 \cos x - 3 \sin x$ prove that $\frac{d^2y}{dx^2}+y = 0$
Answer:
Given function is
$y = 5 \cos x - 3 \sin x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(5\cos x-3\sin x)}{dx}$
$=-5\sin x-3\cos x$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(-5\sin x-3\cos x)}{dx^2}$
$=-5\cos x+3\sin x$
Now,
$\frac{d^2y}{dx^2}+y=-5\cos x+3\sin x+5\cos x-3\sin x = 0$
Hence proved
Question 12: If $y = \cos ^{-1} x$ Find $\frac{d ^2 y }{dx^2 }$ in terms of y alone.
Answer:
Given function is
$y = \cos ^{-1} x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d( \cos ^{-1} x)}{dx}=\frac{-1}{\sqrt{1-x^2}}$
Now, the second order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(\frac{-1}{\sqrt{1-x^2}})}{dx^2}=\frac{-(-1)}{(\sqrt{1-x^2})^2}.(-2x) = \frac{-2x}{1-x^2}$ -(i)
Now, we want $\frac{d^2y}{dx^2}$ in terms of y
$y = \cos ^{-1} x$
$x = \cos y$
Now, put the value of x in (i)
$\frac{d^2y}{dx^2} = \frac{-2\cos y }{1-\cos^2 y } = \frac{-2\cos y}{\sin ^2 y}= -2\cot y cosec y$
$(\because 1-\cos^2x =\sin^2 x\ and \ \frac{\cos x}{\sin x} = \cot x \ and \ \frac{1}{\sin x}= cosec x)$
Therefore, answer is $\frac{d^2y}{dx^2} = -2\cot y cosec y$
Question 13: If $y = 3 \cos (\log x) + 4 \sin (\log x)$ , show that $x^2 y_2 + xy_1 + y = 0$
Answer:
The given function is
$y = 3 \cos (\log x) + 4 \sin (\log x)$
Now, differentiation with respect to. x
$y_1=\frac{dy}{dx}=\frac{d( 3 \cos (\log x) + 4 \sin (\log x))}{dx}=-3\sin(\log x).\frac{1}{x}+4\cos (\log x).\frac{1}{x}$
$=\frac{4\cos (\log x)-3\sin(\log x)}{x}$ -(i)
Now, the second-order derivative is
By using the Quotient rule
$y_2=\frac{d^2y}{dx^2}=\frac{d^2(\frac{4\cos (\log x)-3\sin(\log x)}{x})}{dx^2}= \frac{(-4\sin(\log x).\frac{1}{x}-3\cos(\log x).\frac{1}{x}).x-1.(4\cos (\log x)-3\sin(\log x))}{x^2}$
$=\frac{-\sin(\log x)+7\cos (\log x)}{x^2}$ -(ii)
Now, from equation (i) and (ii), we will get $y_1 \ and \ y_2$
Now, we need to show.
$x^2 y_2 + xy_1 + y = 0$
Put the value of $y_1 \ and \ y_2$ from equation (i) and (ii)
$x^2\left ( \frac{-\sin(\log x)+7\cos (\log x)}{x^2} \right )+x\left ( \frac{4\cos (\log x)-3\sin(\log x)}{x} \right )+ 3 \cos (\log x)$ $+4\sin(\log x)$
$-\sin(\log x)-7\cos(\log x)+4\cos(\log x)-3\sin(\log x)+3\cos (\log x)$ $+4\sin(\log x)$
$=0$
Hence proved
Question 14: If $y = A e ^{mx} + Be ^{nx}$ , show that $\frac{d ^2 y}{dx^2} - (m+n) \frac{dy}{dx} + many = 0$
Answer:
The given function is
$y = A e ^{mx} + Be ^{nx}$
Now, differentiation with respect to. x
$\frac{dy}{dx}=\frac{d(A e ^{mx} + Be ^{nx})}{dx}=mAe^{mx}+nBe^{nx}$ -(i)
Now, the second-order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(mAe^{mx}+nBe^{nx})}{dx^2}= m^2Ae^{mx}+n^2Be^{nx}$ -(ii)
Now, we need to show.
$\frac{d ^2 y}{dx^2} - (m+n) \frac{dy}{dx} + mny = 0$
Put the value of $\frac{d^2y}{dx^2} \ and \ \frac{dy}{dx}$ from equation (i) and (ii)
$m^2Ae^{mx}+n^2Be^{nx}-(m+n)(mAe^{mx}+nBx^{nx}) +mn(Ae^{mx}+Be^{nx})$
$m^2Ae^{mx}+n^2Be^{nx}-m^2Ae^{mx}-mnBx^{nx}-mnAe^{mx} -n^2Be^{nx}+mnAe^{mx}$ $+mnBe^{nx}$
$=0$
Hence proved
Question 15: If $y = 500 e ^{7x} + 600 e ^{- 7x }$ , show that $\frac{d^2 y}{dx ^2} = 49 y$
Answer:
The given function is
$y = 500 e ^{7x} + 600 e ^{- 7x }$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(500 e ^{7x} + 600 e ^{- 7x })}{dx}=7.500e^{7x}-7.600e^{-7x} =3500e^{7x}-4200e^{-7x}$ -(i)
Now, the second-order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(3500e^{7x}-4200e^{-7x})}{dx^2}$
$= 7.3500e^{7x}-(-7).4200e^{-7x}= 24500e^{7x}+29400e^{-7x}$ -(ii)
Now, we need to show.
$\frac{d^2 y}{dx ^2} = 49 y$
Put the value of $\frac{d^2y}{dx^2}$ from equation (ii)
$24500e^{7x}+29400e^{-7x}=49(500e^{7x}+600e^{-7x})$
$= 24500e^{7x}+29400e^{-7x}$
Hence, L.H.S. = R.H.S.
Hence proved
Question 16: If $e ^y (x+1) = 1$ show that $\frac{d^2 y }{dx^2 } = (\frac{dy}{dx})^2$
Answer:
The given function is
$e ^y (x+1) = 1$
We can rewrite it as
$e^y = \frac{1}{x+1}$
Now, differentiation with respect to. x
$\frac{d(e^y)}{dx}=\frac{d(\frac{1}{x+1})}{dx}\\ e^y.\frac{dy}{dx}= \frac{-1}{(x+1)^2}\\ \frac{1}{x+1}.\frac{dy}{dx}= \frac{-1}{(x+1)^2} \ \ \ \ \ \ \ \ \ (\because e^y = \frac{1}{x+1})\\ \frac{dy}{dx}= \frac{-1}{x+1}$ -(i)
Now, the second-order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(\frac{-1}{x+1})}{dx^2}=\frac{-(-1)}{(x+1)^2} = \frac{1}{(x+1)^2}$ -(ii)
Now, we need to show.
$\frac{d^2 y }{dx^2 } = (\frac{dy}{dx})^2$
Put value of $\frac{d^2y}{dx^2} \ and \ \frac{dy}{dx}$ from equation (i) and (ii)
$\frac{1}{(x+1)^2}=\left ( \frac{-1}{x+1} \right )^2$
$=\frac{1}{(x+1)^2}$
Hence, L.H.S. = R.H.S.
Hence proved
Question 17: If $y = (\tan^{-1} x)^2$ show that $(x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2$
Answer:
The given function is
$y = (\tan^{-1} x)^2$
Now, differentiation with respect to. x
$y_1=\frac{dy}{dx}=\frac{d((\tan^{-1}x)^2)}{dx}= 2.\tan^{-1}x.\frac{1}{1+x^2}= \frac{2\tan^{-1}x}{1+x^2}$ -(i)
Now, the second-order derivative is
By using the quotient rule
$y_2=\frac{d^2y}{dx^2}=\frac{d^2(\frac{2\tan^{-1}x}{1+x^2})}{dx^2}=\frac{2.\frac{1}{1+x^2}.(1+x^2)-2\tan^{-1}x(2x)}{(1+x^2)^2}=\frac{2-4x\tan^{-1}x}{(1+x^2)^2}$ -(ii)
Now, we need to show.
$(x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2$
Put the value from equations (i) and (ii)
$(x^2+1)^2.\frac{2-4x\tan^{-1}x}{(1+x^2)^2}+2x(x^2+1).\frac{2\tan^{-1}x}{x^2+1}\\ \rightarrow 2-4x\tan^{-1}x+4x\tan^{-1}x = 2$
Hence, L.H.S. = R.H.S.
Hence proved
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NCERT Continuity and Differentiability Class 12 Solutions: Miscellaneous Exercise
Page number: 144-145
Total questions: 22
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Question 1: Differentiate with respect to. x the function in Exercises 1 to 11.
$( 3x^2 - 9x + 5 )^9$
Answer:
The given function is
$f(x)=( 3x^2 - 9x + 5 )^9$
Now, differentiation w.r.t. x is
$f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d((3x^2-9x+5)^9)}{dx}= 9(3x^2-9x+5)^8.(6x-9)$
$= 27(2x-3)(3x^2-9x+5)^8$
Therefore, differentiation w.r.t. x is $27(3x^2-9x+5)^8(2x-3)$
Question 2: Differentiation with respect to. x the function in Exercises 1 to 11.
Answer:
The given function is
$f(x)= \sin ^3 x + \cos ^6 x$
Now, differentiation w.r.t. x is
$f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d(\sin^3x +\cos^6x)}{dx}=3\sin^2x.\frac{d(\sin x)}{dx}+6\cos^5x.\frac{d(\cos x)}{dx}$
$=3\sin^2x.\cos x+6\cos^5x.(-\sin x)$
$=3\sin^2x\cos x- 6\cos^5x\sin x = 3\sin x\cos x(\sin x- 2\cos ^4x)$
Therefore, differentiation w.r.t. x is $3\sin x\cos x(\sin x- 2\cos ^4x)$
Question 3: Differentiate with respect. x the function in Exercises 1 to 11.
$( 5 x) ^{ 3 \cos 2x }$
Answer:
The given function is
$y=( 5 x) ^{ 3 \cos 2x }$
Take a log on both sides.
$\log y = 3\cos 2x\log 5x$
Now, differentiation w.r.t. x is
By using the product rule
$\frac{1}{y}.\frac{dy}{dx} = 3.(-2\sin 2x)\log 5x + 3\cos 2x.\frac{1}{5x}.5$
$= -6\sin2x\log 5x +\frac{3\cos 2x}{x}\\ \frac{dy}{dx} $
$= y.\left ( -6\sin2x\log 5x +\frac{3\cos 2x}{x} \right )\\$
$ \frac{dy}{dx} = (5x)^{3\cos 2x}.\left ( -6\sin2x\log 5x +\frac{3\cos 2x}{x} \right )$
Therefore, differentiation w.r.t. x is $(5x)^{3\cos 2x}.\left ( \frac{3\cos 2x}{x}-6\sin2x\log 5x \right )$
Question 4: Differentiate with respect to. x the function in Exercises 1 to 11.
$\sin ^ {-1} (x \sqrt x ) , 0 \leq x\leq 1$
Answer:
The given function is
$f(x)=\sin ^ {-1} (x \sqrt x ) , 0 \leq x\leq 1$
Now, differentiation w.r.t. x is
$f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d(\sin^{-1}x\sqrt x)}{dx}=\frac{1}{\sqrt{1-(x\sqrt x)^2}}.\frac{d(x\sqrt x)}{dx}$
$=\frac{1}{\sqrt{1-x^3}}.\left ( 1.\sqrt x+x\frac{1}{2\sqrt x} \right )$
$=\frac{1}{\sqrt{1-x^3}}.\left ( \frac{3\sqrt x}{2} \right )$
$=\frac{3}{2}.\sqrt{\frac{x}{1-x^3}}$
Therefore, differentiation w.r.t. x is $\frac{3}{2}.\sqrt{\frac{x}{1-x^3}}$
Question 5: Differentiate with respect to. x the function in Exercises 1 to 11.
$\frac{\cos ^{-1}x/2}{\sqrt {2x+7}} , -2 < x < 2$
Answer:
The given function is
$f(x)=\frac{\cos ^{-1}x/2}{\sqrt {2x+7}} , -2 < x < 2$
Now, differentiation w.r.t. x is
By using the Quotient rule
$f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d(\frac{\cos^{-1}\frac{x}{2}}{\sqrt{2x+7}})}{dx}$
$=\frac{\frac{d(\cos^{-1}\frac{x}{2})}{dx}.\sqrt{2x+7}-\cos^{-1}\frac{x}{2}.\frac{d(\sqrt{2x+7})}{dx}}{(\sqrt{2x+7})^2}\\$
$ f^{'}(x) = \frac{\frac{-1}{\sqrt{1-(\frac{x}{2})^2}}.\frac{1}{2}.\sqrt{2x+7}-\cos^{-1}\frac{x}{2}.\frac{1}{2.\sqrt{2x+7}}.2}{2x+7}\\$
$ f^{'}(x)= -\left [\frac{1}{(\sqrt{4-x^2})(\sqrt{2x+7})}+\frac{\cos^{-1}\frac{x}{2}}{(2x+7)^\frac{3}{2}} \right ]$
Therefore, differentiation w.r.t. x is $-\left [\frac{1}{(\sqrt{4-x^2})(\sqrt{2x+7})}+\frac{\cos^{-1}\frac{x}{2}}{(2x+7)^\frac{3}{2}} \right ]$
Question 6: Differentiate with respect to. x the function in Exercises 1 to 11.
$\cot ^{-1} \left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} \right ] , 0 < x < \pi /2$
Answer:
The given function is
$f(x)=\cot ^{-1} \left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} \right ] , 0 < x < \pi /2$
Now, rationalise the part.
$\left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} \right ]$
$= \left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} .\frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}\right ]$
$=\frac{(\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x })^2}{(\sqrt{1+\sin x})^2-(\sqrt{1-\sin x})^2}$
$ \ \ \ \ \ \ (Using \ (a-b)(a+b)=a^2-b^2)$
$=\frac{((\sqrt { 1+ \sin x })^2+ (\sqrt { 1- \sin x })^2+2(\sqrt { 1+ \sin x })(\sqrt { 1- \sin x }))}{1+\sin x-1+\sin x}$
$(Using \ (a+b)^2=a^2+b^2+2ab)$
$=\frac{1+\sin x+1-\sin x+2\sqrt{1-\sin^2x} }{2\sin x}$
$=\frac{2(1+\cos x)}{2\sin x} = \frac{1+\cos x}{\sin x}$
$=\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} \ \ \ \ \ (\because 2\cos^2= 1+\cos2x \ and \ \sin2x = 2\sin x\cos x)$
$=\frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} = \cot \frac{x}{2}$
Given function reduces to
$f(x) = \cot^{-1}(\cot \frac{x}{2})\\ f(x) = \frac{x}{2}$
Now, differentiation w.r.t. x is
$f^{'}(x)=\frac{d(f(x))}{dx}$
$=\frac{d(\frac{x}{2})}{dx} = \frac{1}{2}$
Therefore, differentiation w.r.t. x is $\frac{1}{2}$
Question 7: Differentiate with respect to. x the function in Exercises 1 to 11. $( \log x )^{ \log x } , x > 1$
Answer:
The given function is
$y=( \log x )^{ \log x } , x > 1$
Take logs on both sides.
$\log y=\log x\log( \log x )$
Now, differentiate with respect.
$\frac{1}{y}.\frac{dy}{dx}= \frac{1}{x}.\log (\log x)+\log x.\frac{1}{\log x}.\frac{1}{x} = \frac{\log x+1}{x}$
$\frac{dy}{dx} = y.\left ( \frac{\log x+1}{x} \right )\\$
$\frac{dy}{dx} = (\log x)^{\log x}.\left ( \frac{\log x+1}{x} \right )\\$
Therefore, differentiation w.r.t x is $(\log x)^{\log x}.\left ( \frac{\log x+1}{x} \right )\\$
Question 8: $\cos ( \cos x + b \sin x )$, for some constant a and b. r:
The given function is
$f(x)=\cos ( a \cos x + b \sin x )$
Now, differentiation w.r.t x
$f^{'}(x)= \frac{d(f(x))}{dx}= \frac{d(\cos(a\cos x+ b \sin x))}{dx}$
$= -\sin(a\cos x+b\sin x).\frac{d(a\cos x+b\sin x)}{dx}$
$= -\sin(a\cos x+b\sin x).(-a\sin x+b\cos x)$
$= (a\sin x-b\cos x)\sin(a\cos x+b\sin x).$
Therefore, differentiation w.r.t x $(a\sin x-b\cos x)\sin(a\cos x+b\sin x)$
Question 9: $(\sin x - \cos x)^{ (\sin x - \cos x), } , \frac{\pi }{4} <x<\frac{3 \pi }{4}$
Answer:
The given function is
$y=(\sin x - \cos x)^{ (\sin x - \cos x), } , \frac{\pi }{4} <x<\frac{3 \pi }{4}$
Take logs on both sides.
$\log y=(\sin x - \cos x)\log (\sin x - \cos x)$
Now, differentiate with respect. x
$\frac{1}{y}.\frac{dy}{dx} = \frac{d(\sin x-\cos x)}{dx}.\log(\sin x- \cos x)+(\sin x- \cos x).\frac{d(\log(\sin x- \cos x))}{dx}$
$\frac{1}{y}.\frac{dy}{dx} =(\cos x -(-\sin x)).\log(\sin x-\cos x)+(\sin x- \cos x).\frac{(\cos x -(-\sin x))}{(\sin x- \cos x)}$
$\frac{dy}{dx} =y.(\cos x +\sin x)\left ( \log(\sin x-\cos x)+1 \right )$
$\frac{dy}{dx} =(\sin x-\cos x)^{(\sin x-\cos x)}.(\cos x +\sin x)\left ( \log(\sin x-\cos x)+1 \right )$
Therefore, differentiation w.r.t x is $(\sin x-\cos x)^{(\sin x-\cos x)}.(\cos x +\sin x)\left ( \log(\sin x-\cos x)+1 \right ), sinx>cosx$
Question 10: $x ^x + x ^a + a ^x + a ^a$ , for some fixed a > 0 and x > 0
Answer:
The given function is
$f(x)=x ^x + x ^a + a ^x + a ^a$
Let's take
$u = x^x$
Now, take a look at both sides.
$\log u = x \log x$
Now, differentiate w.r.t x
$\frac{1}{u}.\frac{du}{dx}= \frac{dx}{dx}.\log x+x.\frac{d(\log x)}{dx}\\ \\ \frac{1}{u}.\frac{du}{dx}= 1.\log x+x.\frac{1}{x}\\ \\ \frac{du}{dx}= y.(\log x+1)\\ \\ \frac{du}{dx}= x^x.(\log x+1)$ -(i)
Similarly, take $v = x^a$
Take logs on both sides.
$\log v = a\log x$
Now, differentiate w.r.t x
$\frac{1}{v}.\frac{dv}{dx}= a.\frac{d(\log x)}{dx}=a.\frac{1}{x}= \frac{a}{x}\\ \\ \frac{dv}{dx}= v.\frac{a}{x}\\ \\ \frac{dv}{dx}= x^a.\frac{a}{x}$ -(ii)
Similarly, take $z = a^x$
Take logs on both sides.
$\log z = x\log a$
Now, differentiate w.r.t x
$\frac{1}{z}.\frac{dz}{dx}=\log a.\frac{d(x)}{dx}=\log a.1= \log a\\ \\ \frac{dz}{dx}= z.\log a\\ \\ \frac{dz}{dx}= a^x.\log a$ -(iii)
Similarly, take $w = a^a$
Take logs on both sides.
$\log w = a\log a= \ constant$
Now, differentiate w.r.t x
$\frac{1}{w}.\frac{dw}{dx}= a.\frac{d(a\log a)}{dx}= 0\\ \\ \frac{dw}{dx} = 0$ -(iv)
Now,
$f(x)=u+v+z+w$
$f^{'}(x) = \frac{du}{dx}+\frac{dv}{dx}+\frac{dz}{dx}+\frac{dw}{dx}$
Put values from equations (i), (ii),(iii) and (iv)
$f^{'}(x)= x^x(\log x+1)+ax^{a-1}+a^x\log a$
Therefore, differentiation w.r.t. x is $x^x(\log x+1)+ax^{a-1}+a^x\log a$
Question 11: $x ^{x^2 -3} + ( x-3 ) ^{x^2} , for\: \: x > 3$
Answer:
The given function is
$f(x)=x ^{x^2 -3} + ( x-3 ) ^{x^2} , for\: \: x > 3$
take $u=x ^{x^2 -3}$
Now, take a look at both sides.
$\log u=(x^2-3)\log x$
Now, differentiate w.r.t x
$\frac{1}{u}.\frac{du}{dx}= \frac{d(x^2-3)}{dx}.\log x+(x^2-3).\frac{d(\log x)}{dx}\\$
$\frac{1}{u}.\frac{du}{dx} = 2x.\log x+(x^2-3).\frac{1}{x}\\$
$ \frac{1}{u}.\frac{du}{dx} = \frac{2x^2\log x+x^2-3}{x}\\$
$ \frac{du}{dx}= u.\left ( \frac{2x^2\log x+x^2-3}{x} \right )\\$
$ \frac{du}{dx}= x^{(x^2-3)}.\left ( \frac{2x^2\log x+x^2-3}{x} \right )\\$ -(i)
Similarly,
take $v=(x-3)^{x^2}\\$
Now, take a look at both sides.
$\log v=x^2\log (x-3)$
Now, differentiate w.r.t x
$\frac{1}{v}.\frac{dv}{dx}= \frac{d(x^2)}{dx}.\log (x-3)+x^2.\frac{d(\log (x-3))}{dx}\\$
$ \frac{1}{v}.\frac{dv}{dx} = 2x.\log (x-3)+x^2.\frac{1}{(x-3)}\\$
$ \frac{1}{v}.\frac{dv}{dx} = 2x\log(x-3)+\frac{x^2}{x-3}\\$
$ \frac{dv}{dx}= v.\left ( 2x\log(x-3)+\frac{x^2}{x-3} \right )\\$
$ \frac{dv}{dx}= (x-3)^{x^2}.\left ( 2x\log(x-3)+\frac{x^2}{x-3}\right )\\ \\$ -(ii)
Now
$f(x)= u + v$
$f^{'}(x)= \frac{du}{dx}+\frac{dv}{dx}$
Put the value from equations (i) and (ii)
$f^{'}(x)= x^{(x^2-3)}.\left ( \frac{2x^2\log x+x^2-3}{x} \right )+(x-3)^{x^2}.\left ( 2x\log(x-3)+\frac{x^2}{x-3}\right )$
Therefore, differentiation w.r.t x is $x^{(x^2-3)}.\left ( \frac{2x^2\log x+x^2-3}{x} \right )+(x-3)^{x^2}.\left ( 2x\log(x-3)+\frac{x^2}{x-3}\right )$
Question 12: Find $\frac{dy}{dx}$ if $y = 12 (1 - \cos t), x = 10 (t - \sin t),$ $-\frac{\pi }{2} <t< \frac{\pi }{2}$
Answer:
The given equations are
$y = 12 (1 - \cos t), x = 10 (t - \sin t),$
Now, differentiate both y and x w.r.t t independently.
$\frac{dy}{dt}=\frac{d(12(1-\cos t))}{dt}= -12(-\sin t)=12\sin t$
And
$\frac{dx}{dt}=\frac{d(10(t-\sin t))}{dt}= 10-10\cos t$
Now
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{12\sin t}{10(1-\cos t)} = \frac{6}{5}.\frac{2\sin \frac{t}{2}\cos \frac{t}{2}}{2\sin^2\frac{t}{2}} = \frac{6}{5}.\frac{\cos \frac{t}{2}}{\sin \frac{t}{2}}\\ \\$
$(\because \sin 2x = 2\sin x\cos x \ and \ 1-\cos 2x = 2\sin^2x)$
$\frac{dy}{dx}=\frac{6}{5}.\cot \frac{t}{2}$
Therefore, differentiation w.r.t x is $\frac{6}{5}.\cot \frac{t}{2}$
Question 13: Find $\frac{dy}{dx}$ if $y = sin ^{-1} x + sin^{-1} \sqrt{1- x^2} , 0 <x< 1$
Answer:
The given function is
$y = sin ^{-1} x + sin^{-1} \sqrt{1- x^2} , 0 <x< 1$
Now, differentiate with respect. x
$\frac{dy}{dx}= \frac{d(sin ^{-1} x + sin^{-1} \sqrt{1- x^2})}{dx} = \frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-(\sqrt{1-x^2})^2}}.\frac{d(\sqrt{1-x^2})}{dx}\\$
$ \frac{dy}{dx}= \frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-1+x^2}}.\frac{1}{2\sqrt{1-x^2}}.(-2x)\\$
$ \frac{dy}{dx}= \frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}\\$
$ \frac{dy}{dx}= 0$
Therefore, differentiate w.r.t. x is 0
Question 14: If $x \sqrt {1+ y }+ y \sqrt { 1+ x } = 0 \: \: for \: \: , -1 < x < 1 \: \:prove \: \: that \: \frac{dy}{dx} = -\frac{1}{(1+x)^2}$
Answer:
The given function is
$x \sqrt {1+ y }+ y \sqrt { 1+ x } = 0$
$x\sqrt{1+y} = - y\sqrt{1+x}$
Now, squaring both sides.
$(x\sqrt{1+y})^2 = (- y\sqrt{1+x})^2\\ x^2(1+y)=y^2(1+x)\\$
$ x^2+x^2y=y^2x+y^2\\$
$ x^2-y^2=y^2x-x^2y\\ (x-y)(x+y) = -xy(x-y) \\$
$ x+y =-xy\\ y = \frac{-x}{1+x}$
Now, differentiate w.r.t. x is
$\frac{dy}{dx} = \frac{d(\frac{-x}{1+x})}{dx}= \frac{-1.(1+x)-(-x).(1)}{(1+x)^2}= \frac{-1}{(1+x)^2}$
Hence proved
Question 15: If $(x - a)^2 + (y - b)^2 = c^2$ , for some c > 0, prove that $\frac{\left [ 1+(\frac{dy}{dx} )^2 \right ]^{3/2}}{\frac{d^2 y }{dx^2}}\:$ is a constant independent of a and b.
Answer:
The given function is
$(x - a)^2 + (y - b)^2 = c^2$
$(y - b)^2 = c^2-(x - a)^2$ - (i)
Now, differentiate with respect. x
$\frac{d((x-a)^2)}{dx}+\frac{((y-b)^2)}{dx}=\frac{d(c^2)}{dx}\\ \\ 2(x-a)+2(y-b).\frac{dy}{dx}=0\\ \\ \frac{dy}{dx} = \frac{a-x}{y-b}$ -(ii)
Now, the second derivative
$\frac{d^2y}{dx^2} = \frac{\frac{d(a-x)}{dx}.(y-b) -(a-x).\frac{d(y-b)}{dx}}{(y-b)^2}\\ \\ \frac{d^2y}{dx^2} =\frac{ (-1).(y-b)-(a-x).\frac{dy}{dx}}{(y-b)^2}\\ \\$
Now, put values from equations (i) and (ii)
$\frac{d^2y}{dx^2} =\frac{-(y-b)-(a-x).\frac{a-x}{y-b}}{(y-b)^2}\\ \\ \frac{d^2y}{dx^2} = \frac{-((y-b)^2+(a-x)^2)}{(y-b)^\frac{3}{2}} = \frac{-c^2}{(y-b)^\frac{3}{2}}$ $(\because (x - a)^2 + (y - b)^2 = c^2)$
Now,
$\frac{\left [ 1+(\frac{dy}{dx} )^2 \right ]^{3/2}}{\frac{d^2 y }{dx^2}} = \frac{\left ( 1+\left ( \frac{x-a}{y-b} \right )^2 \right )^\frac{3}{2}}{\frac{-c^2}{(y-b)^\frac{3}{2}}} = \frac{\frac{\left ( (y-b)^2 +(x-a)^2\right )^\frac{3}{2}}{(y-b)^\frac{3}{2}}}{\frac{-c^2}{(y-b)^\frac{3}{2}}} = \frac{(c^2)^\frac{3}{2}}{-c^2}= \frac{c^3}{-c^2}= c$ $(\because (x - a)^2 + (y - b)^2 = c^2)$
Which is independent of a and b
Hence proved
Question 16: If $\cos y = x \cos (a + y)$ , with $\cos a \neq \pm 1$ , prove that $\frac{dy}{dx} = \frac{\cos ^2 (a+y )}{\sin a }$
Answer:
The given function is
$\cos y = x \cos (a + y)$
Now, differentiate w.r.t x
$\frac{d(\cos y)}{dx} = \frac{dx}{dx}.\cos(a+y)+x.\frac{d(\cos (a+y))}{dx}\\$
$-\sin y \frac{dy}{dx} = 1.\cos (a+y)+x.(-\sin(a+y)).\frac{dy}{dx}\\$
$ \frac{dy}{dx}.(x\sin(a+y)-\sin y)= \cos(a+y)\\$
$ \frac{dy}{dx}.(\frac{\cos y}{\cos (a+b)}.\sin(a+y)-\sin y)= \cos(a+b) $
$\ \ \ \ \ (\because x = \frac{\cos y}{\cos (a+b)})\\$
$ \frac{dy}{dx}.(\cos y\sin(a+y)-\sin y\cos(a+y))=\cos^2(a+b)\\$
$ \frac{dy}{dx}.(\sin(a+y-y))=\cos^2(a+b)$
$ \ \ \ \ \ \ \ (\because \cos A\sin B-\sin A\cos B = \sin(A-B))\\$
$ \frac{dy}{dx}= \frac{\cos^2(a+y)}{\sin a}$
Hence proved
Question 17: If $x = a (\cos t + t \sin t)$ and $y = a (\sin t - t \cos t),$ find $\frac{d^2 y }{dx^2 }$
Answer:
Given functions are
$x = a (\cos t + t \sin t)$ and $y = a (\sin t - t \cos t)$
Now, differentiate both the functions w.r.t. t independently.
We get
$\frac{dx}{dt} = \frac{d(a(\cos t +t\sin t))}{dt}= a(-\sin t)+a(\sin t+t\cos t)$
$=-a\sin t+a\sin t+at\cos t = at\cos t$
Similarly,
$\frac{dy}{dt} = \frac{d(a(\sin t - t\cos t))}{dt}= a\cos t -a(\cos t+t(-\sin t))$
$= a\cos t -a\cos t+at\sin t =at\sin t$
Now,
$\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{at\sin t}{at \cos t} = \tan t$
Now, the second derivative
$\frac{d^2y}{dx^2}=\frac{d}{dx}\frac{dy}{dx}= \sec^2 t.\frac{dt}{dx}=\frac{\sec^2t.\sec t }{at}=\frac{\sec^3t}{at}$
$(\because \frac{dx}{dt} = at\cos t \rightarrow \frac{dt}{dx}= \frac{1}{at\cos t}=\frac{\sec t}{at})$
Therefore, $\frac{d^2y}{dx^2}=\frac{\sec^3t}{at}$
Question 18: If $f (x) = |x|^3$, show that f ''(x) exists for all real x and find it.
Answer:
The given function is
$f (x) = |x|^3$
$f(x)\left\{\begin{matrix} -x^3 & x<0\\ x^3 & x>0 \end{matrix}\right.$
Now, differentiate in both cases.
$f(x)= x^3\\ f^{'}(x)=3x^2\\ f^{''}(x)= 6x$
And
$f(x)= -x^3\\ f^{'}(x)=-3x^2\\ f^{''}(x)= -6x$
In both cases, f ''(x) exists.
Hence, we can say that f ''(x) exists for all real x
And values are
$f^{''}(x)\left\{\begin{matrix} -6x &x<0 \\ 6x& x>0 \end{matrix}\right.$
Question 19: Using the fact that $\sin (A + B) = \sin A \cos B + \cos A \sin B$ and the differentiation,
Obtain the sum formula for cosines.
Answer:
The given function is
$\sin (A + B) = \sin A \cos B + \cos A \sin B$
Now, differentiate with respect. x
$\frac{d(\sin(A+B))}{dx} = \frac{d\sin A}{dx}.\cos B+\sin A.\frac{d\cos B}{dx}+\frac{d\cos A}{dx}.\sin B+\cos A.\frac{d\sin B}{dx}$
$\cos (A+b)\frac{d(A+B)}{dx}$ $=\frac{dA}{dx}(\cos A\cos B-\sin A\cos B)+\frac{dB}{dx}(\cos A \sin B-\sin A\sin B)$
$=(\cos A \sin B-\sin A\sin B).\frac{d(A+B)}{dx}$
$\cos(A+B)= \cos A\sin B-\sin A\cos B$
Hence, we get the formula by differentiation of sin(A + B)
Question 20: Does there exist a function which is continuous everywhere but not differentiable
At exactly two points? Justify your answer.
Answer:
Consider f(x) = |x| + |x +1|
We know that modulus functions are continuous everywhere, and the sum of two continuous functions is also a continuous function.
Therefore, our function f(x) is continuous.
Now,
If Lets differentiability of our function at x = 0 and x= -1
L.H.D. at x = 0
$\lim\limits_{h\rightarrow 0^-}\frac{f(x+h)-f(x)}{h}= \lim\limits_{h\rightarrow 0^-}\frac{f(h)-f(0)}{h}= \lim\limits_{h\rightarrow 0^-}\frac{|h|+|h+1|-|1|}{h}$
$=\lim\limits_{h\rightarrow 0^-}\frac{-h-(h+1)-1}{h}= 0$ $(|h| = - h \ because\ h < 0)$
R.H.L. at x = 0
$\lim\limits_{h\rightarrow 0^+}\frac{f(x+h)-f(x)}{h}= \lim\limits_{h\rightarrow 0^+}\frac{f(h)-f(0)}{h}= \lim\limits_{h\rightarrow 0^+}\frac{|h|+|h+1|-|1|}{h}$
$=\lim\limits_{h\rightarrow 0^+}\frac{h+h+1-1}{h}= \lim\limits_{h\rightarrow 0^+}\frac{2h}{h}= 2$ $(|h| = h \ because \ h > 0)$
R.H.L. is not equal to L.H.L.
Hence. At x = 0, the function is not differentiable.
Now, Similarly
R.H.L. at x = -1
$\lim\limits_{h\rightarrow 0^+}\frac{f(x+h)-f(x)}{h}= \lim\limits_{h\rightarrow 0^+}\frac{f(-1+h)-f(-1)}{h}= \lim\limits_{h\rightarrow 0^+}\frac{|-1+h|+|h|-|-1|}{h}$
$=\lim\limits_{h\rightarrow 0^+}\frac{1-h+h-1}{h}= \lim\limits_{h\rightarrow 0^+}\frac{0}{h}= 0$ $(|h| = h \ because \ h > 0)$
L.H.L. at x = -1
$\lim\limits_{h\rightarrow 0^-}\frac{f(x+h)-f(x)}{h}= \lim\limits_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h}= \lim\limits_{h\rightarrow 0^-}\frac{|-1+h|+|h|-|1|}{h}$
$=\lim\limits_{h\rightarrow 1^+}\frac{1-h-h-1}{h}= \lim\limits_{h\rightarrow 0^+}\frac{-2h}{h}= -2$ $(|h| = - h \ because\ h < 0)$
L.H.L. is not equal to R.H.L, so not differentiable at x=-1
Hence, exactly two points where it is not differentiable
Question 21: If $y = \begin{vmatrix} f (x) & g(x) & h (x) \\ l& m &n \\ a& b &c \end{vmatrix}$ , prove that $\frac{dy}{dx}$ = $\begin{vmatrix} f '(x) & g'(x) & h' (x) \\ l& m &n \\ a& b &c \end{vmatrix}$
Answer:
Given that
$y = \begin{vmatrix} f (x) & g(x) & h (x) \\ l& m &n \\ a& b &c \end{vmatrix}$
We can rewrite it as
$y = f(x)(mc-bn)-g(x)(lc-an)+h(x)(lb-am)$
Now, differentiate w.r.t x
We will get
$\frac{dy}{dx} = f^{'}(x)(mc-bn)-g^{'}(x)(lc-an)+h^{'}(x)(lb-am) \rightarrow \begin{bmatrix} f^{'}(x) &g^{'}(x) &h^{'}(x) \\ l&m &n \\ a& b &c \end{bmatrix}$
Hence proved
Question 22: If $y=e^{a \cos ^{-1} x},-1 \leq x\leq 1$ , show that $( 1- x^2 ) \frac{d^2 y }{dx ^2} - x \frac{dy}{dx} - a ^2 y = 0$
Answer:
The given function is
$y=e^{a \cos ^{-1} x},-1 \leq 1$
Now, differentiate w.r.t x, we will get.
$\frac{dy}{dx}= \frac{d(e^{a\cos^{-1}x})}{dx}.\frac{d(a\cos^{-1}x)}{dx} = e^{a\cos^{-1}x}.\frac{-a}{\sqrt{1-x^2}}$ -(i)
Now, again differentiate with respect to x
$\frac{d^2y}{dx^2}= \frac{d}{dx}\frac{dy}{dx}= \frac{-ae^{a\cos^{-1}x}.\frac{-a}{\sqrt{1-x^2}}.\sqrt{1-x^2}+ae^{a\cos^{-1}x}.\frac{1.(-2x)}{2\sqrt{1-x^2}}}{(\sqrt{1-x^2})^2}$
$=\ \frac{a^2e^{a\cos^{-1}x}-\frac{axe^{a\cos^{-1}x}}{\sqrt{1-x^2}}}{1-x^2}$ -(ii)
Now, we need to show that.
$( 1- x^2 ) \frac{d^2 y }{dx ^2} - x \frac{dy}{dx} - a ^2 y = 0$
Put the values from equations (i) and (ii)
$(1-x^2).\left ( \ \frac{a^2e^{a\cos^{-1}x}-\frac{axe^{a\cos^{-1}x}}{\sqrt{1-x^2}}}{1-x^2} \right )-x.\left ( \frac{-ae^{a\cos^{-1}x}}{\sqrt{1-x^2}} \right )-a^2e^{a\cos^{-1}x}$
$a^2e^{a\cos^{-1}x}-\frac{axe^{a\cos^{-1}x}}{\sqrt{1-x^2}}+\left ( \frac{axe^{a\cos^{-1}x}}{\sqrt{1-x^2}} \right )-a^2e^{a\cos^{-1}x} = 0$
Hence proved
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