NCERT Solutions for Class 11 Maths Chapter 12 Limits and Derivatives

Limits are considered one of the most important concepts in mathematics, and derivatives are the soul of calculus. Imagine you are climbing a mountain and wonder how steep it is at a certain point. Or, after baking a cake at a high temperature, when you bring it out to room temperature, it will cool down in time, not instantly. You want to know what temperature the cake is at after 30 minutes. Class 11 Maths chapter 12 solutions teach all the relevant concepts for these types of situations, including limits and derivatives. In simpler words, a limit helps us understand the behaviour of a function as it approaches a particular value. On the other hand, a derivative tells us the rate of change of a function. The main purpose of the NCERT class 11 Maths solutions is to strengthen the basics and provide conceptual clarity.

This Story also Contains

1. NCERT Solution for Class 11 Maths Chapter 12 Solutions: Download PDF
2. NCERT Solutions for Class 11 Maths Chapter 12: Exercise Questions
3. Class 11 Maths NCERT Chapter 12: Extra Question
4. Limits and Derivatives Class 11 Chapter 12: Topics
5. Limits and Derivatives Class 11 Solutions: Important Formulae
6. Approach to Solve Questions of Limits and Derivatives Class 11
7. What Extra Should Students Study Beyond NCERT for JEE?
8. NCERT Solutions for Class 11 Mathematics - Chapter Wise

In class 11 Maths NCERT chapter 9 solutions, students will be introduced to calculus, which is an Important branch of mathematics, as well as study algebra of limits and derivatives with their definitions and certain standard functions. Subject matter experts at Careers360 have curated these NCERT class 11 solutions with step-by-step explanations, ensuring clarity in every step. Students can also explore the following links for syllabus, notes, and PDF: NCERT

NCERT Solution for Class 11 Maths Chapter 12 Solutions: Download PDF

Download PDF

NCERT Solutions for Class 11 Maths Chapter 12: Exercise Questions

Question 1: Evaluate the following limits $\lim\limits_{x\rightarrow 3} x +3$

Answer:

$\lim\limits_{x\rightarrow 3} x +3$

$\Rightarrow \lim\limits_{x\rightarrow 3} 3 +3$

$\Rightarrow 6$

Question 2: Evaluate the following limits $\lim\limits_{x \rightarrow \pi } \left ( x - 22/7 \right )$

Answer:

Below, you can find the solution:

$\lim\limits_{x \rightarrow \pi } \left ( x - 22/7 \right )=\pi-\frac{22}{7}$

Question 3: Evaluate the following limits $\lim\limits_{r \rightarrow 1} \pi r^2$

Answer:

The limit

$\lim\limits_{r \rightarrow 1} \pi r^2=\pi(1)^2=\pi$

Hence, the answer is $\pi$

Question 4: Evaluate the following limits $\lim\limits_{x \rightarrow {4}} \frac{4x+3 }{x-2}$

Answer:

The limit

$\lim\limits_{x \rightarrow {4}} \frac{4x+3 }{x-2}$

$\Rightarrow \frac{4(4)+3 }{(4)-2}$

$\Rightarrow \frac{19 }{2}$

Question 5: Evaluate the following limits $\lim\limits_{x \rightarrow {-1}}\frac{x^{10}+ x^5 + 1}{x-1}$

Answer:

The limit

$\lim\limits_{x \rightarrow {4}} \frac{x^{10}+ x^5 + 1 }{x-1}$

$\Rightarrow \frac{(-1)^{10}+ (-1)^5 + 1 }{(-1)-1}$

$\Rightarrow \frac{1-1+1}{-2}$

$\Rightarrow -\frac{1}{2}$

Question 6: Evaluate the following limits $\lim\limits_{x \rightarrow 0 }\frac{( x+1)^5 -1}{x }$

Answer:

The limit

$\lim\limits_{x \rightarrow 0 }\frac{( x+1)^5 -1}{x }$

Let's put

$x+1=y$

Since we have changed the function, its limit will also change,

So

$x\rightarrow 0,y\rightarrow 0+1=1$

So our function has became

$\lim\limits_{y \rightarrow 1 }\frac{ y^5 -1}{y-1 }$

Now, as we know the property

$\lim\limits_{x \rightarrow 1 }\frac{ x^5 -a^n}{x-a }=na^{n-1}$

$\lim\limits_{y \rightarrow 1 }\frac{ y^5 -1}{y-1 }=5(1)^5=5$

Hence,

$\lim\limits_{x \rightarrow 0 }\frac{( x+1)^5 -1}{x }=5$

Question 7: Evaluate the following limits $\lim\limits_{x \rightarrow 2} \frac{3 x^2 - x -10 }{x^2 -4}$

Answer:

The limit

$\lim\limits_{x \rightarrow 2} \frac{3 x^2 - x -10 }{x^2 -4}$

$\Rightarrow \lim\limits_{x \rightarrow 2} \frac{(x-2)(3x+5) }{(x-2)(x+2)}$

$\Rightarrow \lim\limits_{x \rightarrow 2} \frac{(3x+5) }{(x+2)}$

$\Rightarrow \frac{(3(2)+5) }{((2)+2)}$

$\Rightarrow \frac{11 }{4}$

Question 8: Evaluate the following limits $\lim\limits_{x \rightarrow 3} \frac{x ^4 -81}{2x^2 -5x -3}$

Answer:

The limit

$\lim\limits_{x \rightarrow 3} \frac{x ^4 -81}{2x^2 -5x -3}$

At x = 2, both the numerator and denominator become zero, so let's factorise the function

$\lim\limits_{x \rightarrow 3} \frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)}$

$\lim\limits_{x \rightarrow 3} \frac{(x+3)(x^2+9)}{(2x+1)}$

Now we can put the limit directly, so

$\lim\limits_{x \rightarrow 3} \frac{(x+3)(x^2+9)}{(2x+1)}$

$\Rightarrow \frac{((3)+3)((3)^2+9)}{(2(3)+1)}$

$\Rightarrow \frac{6\times18}{7}$

$\Rightarrow \frac{108}{7}$

Question 9: Evaluate the following limits $\lim\limits_{x \rightarrow 0 } \frac{ax +b}{cx+1}$

Answer:

The limit,

$\lim\limits_{x \rightarrow 0 } \frac{ax +b}{cx+1}$

$\Rightarrow \frac{a(0) +b}{c(0)+1}$

$\Rightarrow b$

Question 10: Evaluate the following limits $\lim\limits_{z\rightarrow 1} \frac{z^{1/3}-1}{z^{1/6}-1}$

Answer:

The limit

$\lim\limits_{z\rightarrow 1} \frac{z^{1/3}-1}{z^{1/6}-1}$

Here, on directly putting the limit, both the numerator and the denominator become zero so we factorize the function and then put the limit.

$\lim\limits_{z\rightarrow 1} \frac{z^{1/3}-1}{z^{1/6}-1}=\lim_{z\rightarrow 1} \frac{z^{(1/6)^2}-1^2}{z^{1/6}-1}$

$=\lim\limits_{z\rightarrow 1} \frac{(z^{(1/6)}-1)(z^{(1/6)}+1)}{z^{1/6}-1}$

$=\lim\limits_{z \rightarrow 1}\left(z^{1 / 6}+1\right)$

$=\left(1^{1 / 6}+1\right)$

$=1+1$

$=2$

Question 11: Evaluate the following limits $\lim\limits_{x \rightarrow 1} \frac{ax ^2 +bx + c }{cx^2 + bx + a }, a+b +c \neq 0$

Answer:

The limit:

$\lim\limits_{x \rightarrow 1} \frac{ax ^2 +bx + c }{cx^2 + bx + a }, a+b +c \neq 0$

Since the Denominator is not zero on directly putting the limit, we can directly put the limits, so,

$\lim\limits_{x \rightarrow 1} \frac{ax ^2 +bx + c }{cx^2 + bx + a }, a+b +c \neq 0$

$=\frac{a(1) ^2 +b(1) + c }{c(1)^2 + b(1) + a },$

$=\frac{a+b+c }{a+b+c },$

$=1$

Question 12: Evaluate the following limits $\lim\limits_{x\rightarrow -2} \frac{\frac{1}{x}+ \frac{x}{2}}{x+2}$

Answer:

$\lim\limits_{x\rightarrow -2} \frac{\frac{1}{x}+ \frac{x}{2}}{x+2}$

Here, since the denominator becomes zero on putting the limit directly, we first simplify the function and then put the limit,

$\lim\limits_{x\rightarrow -2} \frac{\frac{1}{x}+ \frac{x}{2}}{x+2}$

$=\lim\limits_{x\rightarrow -2} \frac{\frac{x+2}{2x}}{x+2}$

$=\lim\limits_{x\rightarrow -2} \frac{1}{2x}$

$= \frac{1}{2(-2)}$

$= -\frac{1}{4}$

Question 13: Evaluate the following limits $\lim\limits_{x \rightarrow 0 } \frac{\sin ax }{bx }$

Answer:

The limit

$\lim\limits_{x \rightarrow 0 } \frac{\sin ax }{bx }$

Here, on directly putting the limits, the function becomes $\frac{0}{0}$ form. So we try to make the function in the form of $\frac{sinx}{x}$. so,

$\lim\limits_{x \rightarrow 0 } \frac{\sin ax }{bx }$

$=\lim\limits_{x \rightarrow 0 } \frac{\sin ax(ax) }{bx(ax) }$

$=\lim\limits_{x \rightarrow 0 } \frac{\sin ax }{ax }\frac{a}{b}$

As $\lim\limits_{x\rightarrow 0}\frac{sinx}{x}=1$

$=1\cdot\frac{a}{b}$

$=\frac{a}{b}$

Question 14: Evaluate the following limits $\lim\limits_{x \rightarrow 0} \frac{\sin ax }{\sin bx } , a,b \neq 0$

Answer:

The limit,

$\lim\limits_{x \rightarrow 0} \frac{\sin ax }{\sin bx } , a,b \neq 0$

On putting the limit directly, the function takes the zero by zero form. So, we convert it in the form of $\frac{sina}{a}$ .and then put the limit,

$\Rightarrow \lim\limits_{x\rightarrow {0}}\frac{\frac{sinax}{ax}}{\frac{sinbx}{bx}}\cdot\frac{ax}{bx}$

$=\frac{\lim\limits_{ax\rightarrow {0}}\frac{sinax}{ax}}{\lim\limits_{bx\rightarrow {0}}\frac{sinbx}{bx}}\cdot\frac{a}{b}$

$=\frac{a}{b}$

Question 15: Evaluate the following limits $\lim\limits_{x \rightarrow \pi } \frac{\sin ( \pi -x )}{\pi ( \pi -x)}$

Answer:

The limit

$\lim\limits_{x \rightarrow \pi } \frac{\sin ( \pi -x )}{\pi ( \pi -x)}$

$\Rightarrow \lim\limits_{x \rightarrow \pi } \frac{\sin ( \pi -x )}{ ( \pi -x)}\times\frac{1}{\pi}$

$= 1\times\frac{1}{\pi}$

$= \frac{1}{\pi}$

Question 16: Evaluate the following limits $\lim\limits_{x\rightarrow 0}\frac{\cos x }{\pi -x }$

Answer:

The limit

$\lim\limits_{x\rightarrow 0}\frac{\cos x }{\pi -x }$

The function behaves well on directly putting the limit, so we put the limit directly. So.

$\lim\limits_{x\rightarrow 0}\frac{\cos x }{\pi -x }$

$=\frac{\cos (0) }{\pi -(0) }$

$=\frac{1 }{\pi }$

Question 17: Evaluate the following limits $\lim\limits_{x\rightarrow 0}$ $\frac{\cos 2x -1}{\cos x -1}$

Answer:

The limit:

$\lim\limits_{x\rightarrow 0}$ $\frac{\cos 2x -1}{\cos x -1}$

The function takes the zero-by-zero form when the limit is put directly, so we simplify the function and then put the limit

$\lim\limits_{x\rightarrow 0}$ $\frac{\cos 2x -1}{\cos x -1}$

$\lim\limits_{x\rightarrow 0}$ $\frac{-2(sin^2x)}{-2(sin^2(\frac{x}{2}))}$

$=\lim\limits_{x\rightarrow 0}$ $\frac{\frac{(sin^2x)}{x^2}}{\frac{(sin^2(\frac{x}{2}))}{(\frac{x}{2})^2}}\times\frac{x^2}{(\frac{x}{2})^2}$

$=\frac{1^2}{1^2}\times 4$

$= 4$

Question 18: Evaluate the following limits $\lim\limits_{x \rightarrow 0}\frac{ax + x \cos x }{b \sin x }$

Answer:

$\lim\limits_{x \rightarrow 0}\frac{ax + x \cos x }{b \sin x }$

The function takes the form zero by zero when we put the limit directly in the function Since theunction consists of the sin function and cos function, we try to make the function in the form of $\frac{sinx}{x}$ as we know that it tends to 1 when x tends to 0.

So,

$\lim\limits_{x \rightarrow 0}\frac{ax + x \cos x }{b \sin x }$

$=\frac{1}{b}\lim\limits_{x \rightarrow 0}\frac{x(a+ \cos x) }{ \sin x }$

$=\frac{1}{b}\lim\limits_{x \rightarrow 0}\frac{x }{ \sin x }\times(a+ \cos x)$

$=\frac{1}{b}\times1\times(a+ \cos (0))$

$=\frac{a+1}{b}$

Question 19: Evaluate the following limits $\lim\limits_{x \rightarrow 0} x \sec x$

Answer:

$\lim\limits_{x \rightarrow 0} x \sec x$

As the function doesn't create any abnormality on putting the limit directly, we can put the limit directly. So,

$\lim\limits_{x \rightarrow 0} x \sec x$

$=(0)\times 1$

$=0$ .

Question 20: Evaluate the following limits $\lim\limits_{x\rightarrow 0} \frac{\sin ax + bx }{ax + \sin bx } a,b ,a + b \neq 0$

Answer:

$\lim\limits_{x\rightarrow 0} \frac{\sin ax + bx }{ax + \sin bx } a,b ,a + b \neq 0$

The function takes the zero-by-zero form when we put the limit into the function directly, so we try to eliminate this case by simplifying the function. So

$\lim\limits_{x\rightarrow 0} \frac{\sin ax + bx }{ax + \sin bx } a,b ,a + b \neq 0$

$=\lim\limits_{x\rightarrow 0} \frac{\frac{\sin ax}{ax} \cdot ax+ bx }{ax + \frac{\sin bx}{bx}\cdot bx }$

$=\lim\limits_{x\rightarrow 0} \frac{\frac{\sin ax}{ax} \cdot a+ b }{a + \frac{\sin bx}{bx}\cdot b }$

$=\frac{1\cdot a+b}{a+1\cdot b}$

$=\frac{a+b}{a+ b}$

$=1$

Question 21: Evaluate the following limits $\lim\limits_{x \rightarrow 0} \left ( \csc x - \cot x \right )$

Answer:

$\lim\limits_{x \rightarrow 0} \left ( \csc x - \cot x \right )$

On putting the limit directly, the function takes infinity by infinity form, so we simplify the function and then put the limit

$\lim\limits_{x \rightarrow 0} \left ( \csc x - \cot x \right )$

$=\lim\limits_{x \rightarrow 0} \left (\frac{1}{sinx}-\frac{cosx}{sinx}\right )$

$=\lim\limits_{x \rightarrow 0} \left (\frac{1-cosx}{sinx}\right )$

$=\lim\limits_{x \rightarrow 0} \left (\frac{2sin^2(\frac{x}{2})}{sinx}\right )$

$=\lim\limits_{x \rightarrow 0} \left (\frac{2sin^2(\frac{x}{2})}{(\frac{x}{2})^2}\right )\left ( \frac{(\frac{x}{2})^2}{sinx} \right )$

$=\lim\limits_{x \rightarrow 0} \frac{2}{4}\left (\frac{sin^2(\frac{x}{2})}{(\frac{x}{2})^2}\right )\left ( \frac{(x)}{sinx} \right )\cdot x$

$=\frac{2}{4}\times (1)^2\times0$

$=0$

Question 22: Evaluate the following limits $\lim\limits_{x \rightarrow \pi /2 } \frac{\tan 2x }{x - \pi /2 }$

Answer:

$\lim\limits_{x \rightarrow \pi /2 } \frac{\tan 2x }{x - \pi /2 }$

The function takes zero by zero form when the limit is put directly, so we simplify the function and then put the limits,

So

Let's put

$y=x-\frac{\pi}{2}$

Since we are changing the variable, the limit will also change.

As

$x\rightarrow \frac{\pi}{2},y=x-\frac{\pi}{2}\rightarrow \frac{\pi}{2}-\frac{\pi}{2}=0$

So, the function in the new variable becomes,

$\lim\limits_{y \rightarrow 0 } \frac{\tan 2(y+\frac{\pi}{2}) }{y+\pi/2- \pi /2 }$

$=\lim\limits_{y \rightarrow 0 } \frac{\tan (2y+\pi) }{y }$

As we know that property $tan(\pi+x)=tanx$

$=\lim\limits_{y \rightarrow 0 } \frac{\tan (2y) }{y }$

$=\lim\limits_{y \rightarrow 0 } \frac{sin2y}{2y}\cdot\frac{2}{cos2y}$

$=1\times 2$

$=2$

Question 23: Find $\lim\limits_{x \rightarrow 0} f (x ) \: \: \lim\limits_{x \rightarrow 1} f (x) \: \: where \: \: \: f (x) = \left\{\begin{matrix} 2x+3 & x \leq 0 \\ 3 ( x+1)& x > 0 \end{matrix}\right.$

Answer:

Given Function

$f (x) = \left\{\begin{matrix} 2x+3 & x \leq 0 \\ 3 ( x+1)& x > 0 \end{matrix}\right.$

Now,

Limit at x = 0 :

$at\:x=0^-$

: $\lim\limits_{x\rightarrow{0^-}}f(x)=\lim\limits_{x\rightarrow{0^-}}(2x+3)=2(0)+3=3$

$at\:x=0^+$

$\lim\limits_{x\rightarrow{0^+}}f(x)=\lim\limits_{x\rightarrow{0^+}}3(x+1)=3(0+1)=3$

Hence limit at x = 0 is 3.

Limit at x = 1

$at\:x=1^+$

$\lim\limits_{x\rightarrow{1^+}}f(x)=\lim\limits_{x\rightarrow{1^+}}3(x+1)=3(1+1)=6$

$at\:x=1^-$

$\lim\limits_{x\rightarrow{1^-}}f(x)=\lim\limits_{x\rightarrow{1^-}}3(x+1)=3(1+1)=6$

Hence limit at x = 1 is 6.

Question 24: Find $\lim\limits_{x \rightarrow 1} f (x ) , \: \:where \: \: f (x) = \left\{\begin{matrix} x^2 -1 & x \neq 0 \\ -x^2 -1 & x > 1 \end{matrix}\right.$

Answer:

$\lim\limits_{x \rightarrow 1} f (x ) , \: \:where \: \: f (x) = \left\{\begin{matrix} x^2 -1 & x \neq 0 \\ -x^2 -1 & x > 1 \end{matrix}\right.$

Limit at $x=1^+$

$\lim\limits_{x \rightarrow 1^+} f (x ) = \lim\limits_{x \rightarrow 1} (-x^2-1)=-(1)^2-1=-2$

Limit at $x=1^-$

$\lim\limits_{x \rightarrow 1^-} f (x ) = \lim\limits_{x \rightarrow 1} (x^2-1)=(1)^2-1=0$

As we can see Limit at $x=1^+$ is not equal to the Limit at $x=1^-$, The limit of this function at x = 1 does not exist.

Question 25: Evaluate $\lim\limits_{x \rightarrow 0} f (x) , \: \: where \: \: f (x) = \left\{\begin{matrix} \frac{|x|}{x} & x \neq 0 \\ 0 & x = 0 \end{matrix}\right.$

Answer:

$\lim\limits_{x \rightarrow 0} f (x) , \: \: where \: \: f (x) = \left\{\begin{matrix} \frac{|x|}{x} & x \neq 0 \\ 0 & x = 0 \end{matrix}\right.$

The right-hand Limit or Limit at $x=0^+$

$\lim\limits_{x \rightarrow 0^+} f (x) = \lim\limits_{x \rightarrow 0^+} \frac{|x|}{x}=\frac{x}{x}=1$

The left-hand limit or Limit at $x=0^-$

$\lim\limits_{x \rightarrow 0^-} f (x) = \lim\limits_{x \rightarrow 0^-} \frac{|x|}{x}=\frac{-x}{x}=-1$

Since The left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exist.

Question 26: Evaluate $\lim\limits_{x \rightarrow 0} f (x) , \: \: where \: \: f (x) = \left\{\begin{matrix} \frac{x}{|x|} & x \neq 0 \\ 0 & x = 0 \end{matrix}\right.$

Answer:

$\lim\limits_{x \rightarrow 0} f (x) , \: \: where \: \: f (x) = \left\{\begin{matrix} \frac{x}{|x|} & x \neq 0 \\ 0 & x = 0 \end{matrix}\right.$

The right-hand Limit or Limit at $x=0^+$

$\lim\limits_{x \rightarrow 0^+} f (x) = \lim\limits_{x \rightarrow 0^+} \frac{x}{|x|}=\frac{x}{x}=1$

The left-hand limit or Limit at $x=0^-$

$\lim\limits_{x \rightarrow 0^-} f (x) = \lim\limits_{x \rightarrow 0^-} \frac{x}{|x|}=\frac{x}{-x}=-1$

Since The left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exist.

Question 27: Find $\lim\limits_{x \rightarrow 5 } f (x) , where f( x) = |x| -5$

Answer:

$\lim\limits_{x \rightarrow 5 } f (x) , where f( x) = |x| -5$

The right-hand Limit or Limit at $x=5^+$

$\lim\limits_{x \rightarrow 5^+} f (x) = \lim\limits_{x \rightarrow 5^+} |x|-5=5-5=0$

The left-hand limit or Limit at $x=5^-$

$\lim\limits_{x \rightarrow 5^-} f (x) = \lim\limits_{x \rightarrow 5^-}|x|-5=5-5=0$

Since The left-hand limit and right-hand limit are equal, the limit of this function at x = 5 is 0.

Question 28: Suppose $f (x) = \left\{\begin{matrix} a+bx & x < 1 \\ 4 & x = 1 \\ b - ax & x > 1 \end{matrix}\right.$ f (x) = f (1) what are possible values of a and b?

Answer:

Given,

$f (x) = \left\{\begin{matrix} a+bx & x < 1 \\ 4 & x = 1 \\ b - ax & x > 1 \end{matrix}\right.$

And

$\lim\limits_{x\rightarrow 1} f(x)=f(1)$

Since the limit exists,

left-hand limit = Right-hand limit = f(1).

Left-hand limit = f(1)

$\lim\limits_{x\rightarrow 1^-} f(x)= \lim_{x\rightarrow 1}(a+bx)=a+b(1)=a+b=4$

Right-hand limit

$\lim\limits_{x\rightarrow 1^+} f(x)= \lim\limits_{x\rightarrow 1}(b-ax)=b-a(1)=b-a=4$

From both equations, we get that,

$a=0$ and $b=4$

Hence, the possible values of a and b are 0 and 4, respectively.

Question 29: Let a1, a2, . . ., an be fixed real numbers and define a function $f (x) = (x - a_1 ) (x - a_2 )...(x - a_n ) .$ What is $\lim\limits_{x \rightarrow a _ 1 }$ f (x) ? For some $a \neq a _ 1 , a _ 2 .... a _n$ , compute l $\lim\limits_{ x \rightarrow a } f (x)$

Answer:

Given,

$f (x) = (x - a_1 ) (x - a_2 )...(x - a_n ) .$

Now,

$\\\lim\limits_{x \rightarrow a _ 1 }f(x)=\lim\limits_{x \rightarrow a _ 1 }[(x - a_1 ) (x - a_2 )...(x - a_n ) ]\\.=[\lim\limits_{x \rightarrow a _ 1 }(x - a_1 )][\lim\limits_{x \rightarrow a _ 1 }(x - a_2 )][\lim\limits_{x \rightarrow a _ 1 }(x - a_n )] \\=0$

Hence,

$\lim\limits_{x \rightarrow a _ 1 }f(x)=0$

Now,

$\lim\limits_{ x \rightarrow a } f (x)=\lim\limits_{ x \rightarrow a } (x-a_1)(x-a_2)...(x-a_n)$

$\lim\limits_{ x \rightarrow a } f (x)=(a-a_1)(a-a_2)(a-a_3)$

Hence

$\lim\limits_{ x \rightarrow a } f (x)=(a-a_1)(a-a_2)(a-a_3)$ .

Question 30: If $f (x) = \left\{\begin{matrix} |x| + 1 & x < 0 \\ 0 & x = 0 \\ |x| -1& x > 0 \end{matrix}\right.$ For what value (s) of a does $\lim\limits_{x \rightarrow a } f (x)$ exists ?

Answer:

$f (x) = \left\{\begin{matrix} |x| + 1 & x < 0 \\ 0 & x = 0 \\ |x| -1& x > 0 \end{matrix}\right.$

The limit at x = a exists when the right-hand limit is equal to the left-hand limit. So,

Case 1: when a = 0

The right-hand Limit or Limit at $x=0^+$

$\lim\limits_{x \rightarrow 0^+} f (x) = \lim\limits_{x \rightarrow 0^+} |x|-1=1-1=0$

The left-hand limit or Limit at $x=0^-$

$\lim\limits_{x \rightarrow 0^-} f (x) = \lim\limits_{x \rightarrow 0^-} |x|+1=0+1=1$

Since the Left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exist.

Case 2: When a < 0

The right-hand Limit or Limit at $x=a^+$

$\lim\limits_{x \rightarrow a^+} f (x) = \lim\limits_{x \rightarrow a^+} |x|-1=a-1$

The left-hand limit or Limit at $x=a^-$

$\lim\limits_{x \rightarrow a^-} f (x) = \lim\limits_{x \rightarrow a^-} |x|-1=a-1$

Since LHL = RHL, the Limit exists at x = a and is equal to a-1.

Case 3: When a > 0

The right-hand Limit or Limit at $x=a^+$

$\lim\limits_{x \rightarrow a^+} f (x) = \lim\limits_{x \rightarrow a^+} |x|+1=a+1$

The left-hand limit or Limit at $x=a^-$

$\lim\limits_{x \rightarrow a^-} f (x) = \lim\limits_{x \rightarrow a^-} |x|+1=a+1$

Since LHL = RHL, Limit exists at x = a and is equal to a+1

Hence,

The limit exists at all points except at x=0.

Question 31:If the function f(x) satisfies $\lim\limits_{x \rightarrow 1} \frac{f (x)-2}{x^2-1} = \pi$ , evaluate $\lim\limits_{x \rightarrow 1} f (x)$

Answer:

Given

$\lim\limits_{x \rightarrow 1} \frac{f (x)-2}{x^2-1} = \pi$

Now,

$\lim\limits_{x \rightarrow 1} \frac{f (x)-2}{x^2-1} = \frac{\lim\limits_{x \rightarrow 1}(f (x)-2)}{\lim\limits_{x \rightarrow 1}(x^2-1)}=\pi$

${\lim\limits_{x \rightarrow 1}(f (x)-2)}=\pi{\lim\limits_{x \rightarrow 1}(x^2-1)}$

${\lim\limits_{x \rightarrow 1}(f (x)-2)}=\pi{(1-1)}$

${\lim\limits_{x \rightarrow 1}(f (x)-2)}=0$

${\lim\limits_{x \rightarrow 1}f (x)}=2$

Question 32: If $f(x)=\left[\begin{array}{cl}m x^2+n & ; x<0 \\ n x+m & ; x \leq 0 \leq 1 t \\ n x^3+m & ; x>1\end{array}\right]$

Answer:

Given,

$f(x)=\left[\begin{array}{cl}m x^2+n & ; x<0 \\ n x+m & ; x \leq 0 \leq 1 t \\ n x^3+m & ; x>1\end{array}\right]$

Case 1: Limit at x = 0

The right-hand Limit or Limit at $x=0^+$

$\lim\limits_{x \rightarrow 0^+} f (x) = \lim\limits_{x \rightarrow 0^+} nx+m=n(0)+m=m$

The left-hand limit or Limit at $x=0^-$

$\lim\limits_{x \rightarrow 0^-} f (x) = \lim\limits_{x \rightarrow 0^-} mx^2+n=m(0)^2+n=n$

Hence Limit will exist at x = 0 when m = n.

Case 2: Limit at x = 1

The right-hand Limit or Limit at $x=1^+$

$\lim\limits_{x \rightarrow 1^+} f (x) = \lim\limits_{x \rightarrow 1^+} nx^3+m=n(1)^3+m=n+m$

The left-hand limit or Limit at $x=1^-$

$\lim\limits_{x \rightarrow 1^-} f (x) = \lim\limits_{x \rightarrow 1^-} nx+m=n(1)+m=n+m$

Hence Limit at 1 exists at all integers.

Question 1: Find the derivative of $x ^ 2 -2 \: \: at \: \: x = 10$

Answer:

F(x)= $x ^ 2 -2 \: \:$

Now, As we know, The derivative of any function at x is

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

The derivative of f(x) at x = 10:

$f'(10)=\lim\limits_{h\rightarrow 0}\frac{f(10+h)-f(10)}{h}$

$f'(10)=\lim\limits_{h\rightarrow 0}\frac{(10+h)^2-2-((10)^2-2)}{h}$

$f'(10)=\lim\limits_{h\rightarrow 0}\frac{100+20h+h^2-2-100 +2}{h}$

$f'(10)=\lim\limits_{h\rightarrow 0}\frac{20h+h^2}{h}$

$f'(10)=\lim\limits_{h\rightarrow 0}20+h$

$f'(10)=20+0$

$f'(10)=20$

Question 2: Find the derivative of x at x = 1.

Answer:

Given

f(x)= x

Now, As we know, The derivative of any function at x is

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

The derivative of f(x) at x = 1:

$f'(1)=\lim\limits_{h\rightarrow 0}\frac{f(1+h)-f(1)}{h}$

$f'(1)=\lim\limits_{h\rightarrow 0}\frac{(1+h)-(1)}{h}$

$f'(1)=\lim\limits_{h\rightarrow 0}\frac{(h)}{h}$

$f'(1)=1$ (Answer)

Question 3: Find the derivative of 99x at x = l00.

Answer:

f(x)= 99x

Now, as we know, the derivative of any function at x is

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

The derivative of f(x) at x = 100:

$f'(100)=\lim\limits_{h\rightarrow 0}\frac{f(100+h)-f(100)}{h}$

$f'(100)=\lim\limits_{h\rightarrow 0}\frac{99(100+h)-99(100)}{h}$

$f'(100)=\lim\limits_{h\rightarrow 0}\frac{99h}{h}$

$f'(100)=99$

Question 4 (i): Find the derivative of the following functions from the first principle. $x ^3 -27$

Answer:

Given

f(x)= $x ^3 -27$

Now, as we know, the derivative of any function at x is

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{(x+h)^3-27-((x)^3-27)}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{x^3+h^3+3x^2h+3h^2x-27+x^3+27}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{h^3+3x^2h+3h^2x}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}{h^2+3x^2+3hx}$

$f'(x)=3x^2$

Question 4(ii): Find the derivative of the following function from the first principle. $( x-1)(x-2)$

Answer:

f(x)= $( x-1)(x-2)$

Now, as we know, the derivative of any function at x is

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{(x+h-1)(x+h-2)-(x-1)(x-2)}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{x^2+xh-2x+hx+h^2-2h-x-h+2-x^2+2x+x-2}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{2hx+h^2-3h}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}{2x+h-3}$

$f'(x)=2x-3$ (Answer)

Question 4 (iii): Find the derivative of the following functions from the first principle. $1 / x ^2$

Answer:

f(x)= $1 / x ^2$

Now, as we know, the derivative of any function at x is

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{1/(x+h)^2-1/(x^2)}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0} \frac{\frac{x^2-(x+h)^2}{(x+h)^2x^2}}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0} \frac{x^2-x^2-2xh-h^2}{h(x+h)^2x^2}$

$f'(x)=\lim\limits_{h\rightarrow 0} \frac{-2xh-h^2}{h(x+h)^2x^2}$

$f'(x)=\lim\limits_{h\rightarrow 0} \frac{-2x-h}{(x+h)^2x^2}$

$f'(x)= \frac{-2x-0}{(x+0)^2x^2}$

$f'(x)= \frac{-2}{x^3}$ (Answer)

Question 4 (iv): Find the derivative of the following functions from the first principle. $\frac{x +1}{x-1}$

Answer:

Given:

$f(x)=\frac{x +1}{x-1}$

Now, as we know, the derivative of any function at x is

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{\frac{x+h+1}{x+h-1}-\frac{x+1}{x-1}}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{\frac{(x+h+1)(x-1)-(x+1)(x+h-1)}{(x-1)(x+h-1)}}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{x^2-x+hx-h+x-1-x^2-xh+x-x-h+1}{(x-1)(x+h-1)h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{-2h}{(x-1)(x+h-1)h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{-2}{(x-1)(x+h-1)}$

$f'(x)=\frac{-2}{(x-1)(x+0-1)}$

$f'(x)=\frac{-2}{(x-1)^2}$

Question 5: For the function $f (x) = \frac{x^{100}}{100}+ \frac{x ^ {99}}{99} + ....+ \frac{x ^2 }{2}+ x+ 1$ Prove that f '(1) =100 f '(0).

Answer:

$f (x) = \frac{x^{100}}{100}+ \frac{x ^ {99}}{99} + ....+ \frac{x ^2 }{2}+ x+ 1$

As we know, the property,

$f'(x^n)=nx^{n-1}$

Applying that property, we get

$f '(x) = 100\frac{x^{99}}{100}+ 99\frac{x ^ {98}}{99} + ....+ 2\frac{x }{2}+ 1+ 0$

$f '(x) = x^{99}+x^{98}+......x+1$

Now.

$f '(0) = 0^{99}+0^{98}+......0+1=1$

$f '(1) = 1^{99}+1^{98}+......1+1=100$

So,

Here

$1\times 100=100$

$f'(0)\times 100=f'(1)$

Hence Proved.

Question 6: Find the derivative of $x ^n + ax ^{n-1} + a ^ 2 x ^{n-2} + ....+ a ^ { n-1} x + a ^n$ for some fixed real number a.

Answer:

Given

$f(x)=x ^n + ax ^{n-1} + a ^ 2 x ^{n-2} + ....+ a ^ { n-1} x + a ^n$

As we know, the property,

$f'(x^n)=nx^{n-1}$

Applying that property, we get

$f'(x)=nx ^{n-1} + a(n-1)x ^{n-2} + a ^ 2(n-2) x ^{n-3} + ....+ a ^ { n-1} 1 + 0$

$f'(x)=nx ^{n-1} + a(n-1)x ^{n-2} + a ^ 2(n-2) x ^{n-3} + ....+ a ^ { n-1}$

Question 7(i): For some constants a and b, find the derivative of $( x - a ) ( x -b )$

Answer:

Given

$f(x)=( x - a ) ( x -b )=x^2-ax-bx+ab$

As we know, the property,

$f'(x^n)=nx^{n-1}$

And the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying that property, we get

$f'(x)=2x-a-b$

Question 7(ii): For some constants a and b, find the derivative of $( ax ^2 + b)^2$

Answer:

Given

$f(x)=( ax ^2 + b)^2=a^2x^4+2abx^2+b^2$

As we know, the property,

$f'(x^n)=nx^{n-1}$

And the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying those properties, we get

$f'(x)=4a^2x^3+2(2)abx+0$

$f'(x)=4a^2x^3+4abx$

$f'(x)=4ax(ax^2+b)$

Question 7(iii): For some constants a and b, find the derivative of $\frac{x - a }{x -b }$

Answer:

Given,

$f(x)=\frac{x - a }{x -b }$

Now, as we know the quotient rule of derivative,

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{dy_1}{dx}-y_1\frac{dy_2}{dx}}{y_2^2}$

So,, applying this rule, we get

$\frac{d(\frac{x-a}{x-b})}{dx}=\frac{(x-b)\frac{d(x-a)}{dx}-(x-a)\frac{d(x-b)}{dx}}{(x-b)^2}$

$\frac{d(\frac{x-a}{x-b})}{dx}=\frac{(x-b)-(x-a)}{(x-b)^2}$

$\frac{d(\frac{x-a}{x-b})}{dx}=\frac{a-b}{(x-b)^2}$

Hence

$f'(x)=\frac{a-b}{(x-b)^2}$

Question 8: Find the derivative of $\frac{x ^n - a ^n }{x - a }$ for some constant a.

Answer:

Given,

$f(x)=\frac{x ^n - a ^n }{x - a }$

Now, as we know the quotient rule of derivative,

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{dy_1}{dx}-y_1\frac{dy_2}{dx}}{y_2^2}$

So, applying this rule, we get

$\frac{d(\frac{x^n-a^n}{x-a})}{dx}=\frac{(x-a)\frac{d(x^n-a^n)}{dx}-(x^n-a^n)\frac{d(x-a)}{dx}}{(x-a)^2}$

$\frac{d(\frac{x^n-a^n}{x-a})}{dx}=\frac{(x-a)nx^{n-1}-(x^n-a^n)}{(x-a)^2}$

$\frac{d(\frac{x^n-a^n}{x-a})}{dx}=\frac{nx^n-anx^{n-1}-x^n+a^n}{(x-a)^2}$

Hence

$f'(x)=\frac{nx^n-anx^{n-1}-x^n+a^n}{(x-a)^2}$

Question 9(i): Find the derivative of $2x - 3/4$

Answer:

Given:

$f(x)=2x - 3/4$

As we know, the property,

$f'(x^n)=nx^{n-1}$

And the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying that property, we get

$f'(x)=2-0$

$f'(x)=2$

Question 9(ii): Find the derivative of $( 5x^3 + 3x -1 ) ( x -1)$

Answer:

Given.

$f(x)=( 5x^3 + 3x -1 ) ( x -1)=5x^4+3x^2-x-5x^3-3x+1$

$f(x)=5x^4-5x^3+3x^2-4x+1$

As we know, the property,

$f'(x^n)=nx^{n-1}$

And the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying that property, we get

$f'(x)=5(4)x^3-5(3)x^2+3(2)x-4+0$

$f'(x)=20x^3-15x^2+6x-4$

Question 9(iii): Find the derivative of $x ^{-3} ( 5 + 3x )$

Answer:

Given

$f(x)=x ^{-3} ( 5 + 3x )=5x^{-3}+3x^{-2}$

As we know, the property,

$f'(x^n)=nx^{n-1}$

And the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying that property, we get

$f'(x)=(-3)5x^{-4}+3(-2)x^{-3}$

$f'(x)=-15x^{-4}-6x^{-3}$

Question 9(iv): Find the derivative of $x ^5 ( 3 - 6 x ^{-9})$

Answer:

Given

$f(x)=x ^5 ( 3 - 6 x ^{-9})=3x^5-6x^{-4}$

As we know, the property,

$f'(x^n)=nx^{n-1}$

And the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying that property, we get

$f'(x)=(5)3x^4-6(-4)x^{-5}$

$f'(x)=15x^4+24x^{-5}$

Question 9(v): Find the derivative of $x ^{-4} ( 3 - 4x ^{-5})$

Answer:

Given

$f(x)=x ^{-4} ( 3 - 4x ^{-5})=3x^{-4}-4x^{-9}$

As we know, the property,

$f'(x^n)=nx^{n-1}$

And the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying that property, we get

$f'(x)=(-4)3x^{-5}-(-9)4x^{-10}$

$f'(x)=-12x^{-5}+36x^{-10}$

Question 9(vi): Find the derivative of $\frac{2}{x+1}- \frac{x^2 }{3 x-1}$

Answer:

Given

$f(x)=\frac{2}{x+1}- \frac{x^2 }{3 x-1}$

As we know, the quotient rule of the derivative is:

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{dy_1}{dx}-y_1\frac{dy_2}{dx}}{y_2^2}$

And the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

So, applying this rule, we get

$\frac{d(\frac{2}{x+1}-\frac{x^2}{3x-1})}{dx}=\frac{(x+1)\frac{d(2)}{dx}-2\frac{d(x+1)}{dx}}{(x+1)^2}-\frac{(3x-1)\frac{d(x^2)}{dx}-x^2\frac{d(3x-1)}{dx}}{(3x-1)^2}$

$\frac{d(\frac{2}{x+1}-\frac{x^2}{3x-1})}{dx}=\frac{-2}{(x+1)^2}-\frac{(3x-1)2x-x^23}{(3x-1)^2}$

$\frac{d(\frac{2}{x+1}-\frac{x^2}{3x-1})}{dx}=\frac{-2}{(x+1)^2}-\frac{6x^2-2x-3x^2}{(3x-1)^2}$

$\frac{d(\frac{2}{x+1}-\frac{x^2}{3x-1})}{dx}=\frac{-2}{(x+1)^2}-\frac{3x^2-2x}{(3x-1)^2}$

Hence

$f'(x)=\frac{-2}{(x+1)^2}-\frac{3x^2-2x}{(3x-1)^2}$

Question 10: Find the derivative of $\cos x$ from the first principle.

Answer:

Given,

f(x)= $\cos x$

Now, as we know, the derivative of any function at x is

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{\cos(x+h)-\cos(x)}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{\cos(x)\cos(h)-\sin(x)\sin(h)-\cos(x)}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{\cos(x)\cos(h)-\cos(x)}{h}-\frac{\sin(x)\sin(h)}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{\cos(x)(\cos(h)-1)}{h}-\frac{\sin(x)\sin(h)}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\cos(x)\frac{-2sin^2(h/2)}{h}\cdot -\sin(x)\frac{sinh}{h}$

$f'(x)=\cos(x)(0)-sinx(1)$

$f'(x)=-\sin(x)$

Question 11(i): Find the derivative of the following functions: $\sin x \cos x$

Answer:

Given,

f(x)= $\sin x \cos x$

Now, as we know the product rule of derivative,

$\frac{d(y_1y_2)}{dx}=y_1\frac{dy_2}{dx}+y_2\frac{dy_1}{dx}$

So, applying the rule here,

$\frac{d(\sin x\cos x)}{dx}=\sin x\frac{d\cos x}{dx}+\cos x\frac{d\sin x}{dx}$

$\frac{d(\sin x\cos x)}{dx}=\sin x(-\sin x)+\cos x (\cos x)$

$\frac{d(\sin x\cos x)}{dx}=-\sin^2 x+\cos^2 x$

$\frac{d(\sin x\cos x)}{dx}=\cos 2x$

Question 11(ii): Find the derivative of the following functions: $\sec x$

Answer:

Given

$f(x)=\sec x=\frac{1}{\cos x}$

Now, as we know the quotient rule of derivative,

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{dy_1}{dx}-y_1\frac{dy_2}{dx}}{y_2^2}$

So, applying this rule, we get

$\frac{d(\frac{1}{\cos x})}{dx}=\frac{\cos x\frac{d(1)}{dx}-1\frac{d(\cos x)}{dx}}{\cos ^2x}$

$\frac{d(\frac{1}{\cos x})}{dx}=\frac{-1(-\sin x)}{\cos ^2x}$

$\frac{d(\frac{1}{\cos x})}{dx}=\frac{\sin x}{\cos ^2x}=\frac{\sin x}{\cos x}\frac{1}{\cos x}$

$\frac{d(\sec x)}{dx}=\tan x\sec x$

Question 11 (iii): Find the derivative of the following functions: $5 \sec x + 4 \cos x$

Answer:

Given

$f(x)=5 \sec x + 4 \cos x$

As we know, the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying the property, we get

$\frac{d(5\sec x+4\cos x)}{dx}=\frac{d(5\sec x)}{dx}+\frac{d(4\cos x)}{dx}$

$\frac{d(5\sec x+4\cos x)}{dx}=5\tan x\sec x+4(-\sin x)$

$\frac{d(5\sec x+4\cos x)}{dx}=5\tan x\sec x-4\sin x$

Question 11(iv): Find the derivative of the following functions: $\csc x$

Answer:

Given :

$f(x)=\csc x=\frac{1}{\sin x}$

Now, as we know the quotient rule of derivative,

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{dy_1}{dx}-y_1\frac{dy_2}{dx}}{y_2^2}$

So, applying this rule, we get

$\frac{d(\frac{1}{\sin x})}{dx}=\frac{(\sin x)\frac{d(1)}{dx}-1\frac{d(\sin x)}{dx}}{(\sin x)^2}$

$\frac{d(\frac{1}{\sin x})}{dx}=\frac{-1(\cos x)}{(\sin x)^2}$

$\frac{d(\frac{1}{\sin x})}{dx}=-\frac{(\cos x)}{(\sin x)}\frac{1}{\sin x}$

$\frac{d(\csc x)}{dx}=-\cot x \csc x$

Question 11(v): Find the derivative of the following functions: $3 \cot x + 5 \csc x$

Answer:

Given,

$f(x)=3 \cot x + 5 \csc x$

As we know, the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying the property,

$\frac{d(3\cot x+5 \csc x)}{dx}=\frac{d(3\cot x)}{dx}+\frac{d(5\csc x)}{dx}$

$\frac{d(3\cot x+5 \csc x)}{dx}=3\frac{d(\frac{\cos x}{\sin x})}{dx}+\frac{d(5\csc x)}{dx}$

$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x+3\frac{d(\frac{\cos x}{\sin x})}{dx}$

Now, as we know the quotient rule of the derivative,

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{dy_1}{dx}-y_1\frac{dy_2}{dx}}{y_2^2}$

So, applying this rule, we get

$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x+3\left[\frac{\sin x\frac{d(\cos x)}{dx}-\cos x(\frac{d(\sin x)}{dx})}{\sin^2x}\right]$

$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x+3\left[\frac{\sin x(-\sin x)-\cos x(\cos x)}{\sin^2x}\right]$

$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x+3\left[\frac{-\sin^2 x-\cos^2 x}{\sin^2x}\right]$

$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x-3\left[\frac{\sin^2 x+\cos^2 x}{\sin^2x}\right]$

$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x-3\left[\frac{1}{\sin^2x}\right]$

$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x-3\csc^2x$

Question 11(vi): Find the derivative of the following functions: $5 \sin x - 6 \cos x + 7$

Answer:

Given,

$f(x)=5 \sin x - 6 \cos x + 7$

Now, as we know the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

So, applying the property,

$f'(x)=5 \cos x - 6 (-\sin x ) + 0$

$f'(x)=5 \cos x + 6 (\sin x )$

$f'(x)=5 \cos x + 6 \sin x$

Question 11(vii): Find the derivative of the following functions: $2 \tan x - 7 \sec x$

Answer:

Given

$f(x)=2 \tan x - 7 \sec x$

As we know, the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying this property,

$\frac{d(2\tan x+7\sec x)}{dx}=2\frac{d\tan x}{dx}+7\frac{d\sec x}{dx}$

$\frac{d(2\tan x+7\sec x)}{dx}=2\sec^2x +7(-\sec x\tan x)$

$\frac{d(2\tan x+7\sec x)}{dx}=2\sec^2x -7\sec x\tan x$

Question 1(i): Find the derivative of the following functions from the first principle: -x

Answer:

Given.

f(x)=-x

Now, as we know, the derivative of any function at x is

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{-(x+h)-(-x)}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{-h}{h}$

$f'(x)=-1$

Question 1 (ii): Find the derivative of the following functions from the first principle: $( - x ) ^{-1}$

Answer:

Given.

f(x)= $( - x ) ^{-1}$

Now, as we know, the derivative of any function at x is

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{-(x+h)^{-1}-(-x)^{-1}}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{-\frac{1}{x+h}+\frac{1}{x}}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{\frac{-x+x+h}{x(x+h)}}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{\frac{h}{(x+h)(x)}}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{1}{x(x+h)}$

$f'(x)=\frac{1}{x^2}$

Question 1 (iii): Find the derivative of the following functions from the first principle: $\sin ( x+1)$

Answer:

Given.

$f(x)=\sin ( x+1)$

Now, as we know, the derivative of any function at x is

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{\sin(x+h+1)-\sin(x+1)}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{2\cos(\frac{x+h+1+x+1}{2})\sin(\frac{x+h+1-x-1}{2})}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{2\cos(\frac{2x+h+2}{2})\sin(\frac{h}{2})}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{\cos(\frac{2x+h+2}{2})\sin(\frac{h}{2})}{\frac{h}{2}}$

$f'(x)=\cos(\frac{2x+0+2}{2})\times 1$

$f'(x)=\cos(x+1)$

Question 1(iv): Find the derivative of the following functions from the first principle: $\cos ( x - \pi /8 )$

Answer:

Given.

$f(x)=\cos ( x - \pi /8 )$

Now, as we know, the derivative of any function at x is

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{\cos(x+h-\pi/8)-cos(x-\pi/8)}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{-2\sin\left ( \frac{x+h-\pi/8+x-\pi/8 }{2}\right )\sin\left ( \frac{x+h-\pi/8-x+\pi/8}{2} \right )}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{-2\sin\left ( \frac{2x+h-\pi/4 }{2}\right )\sin\left ( \frac{h}{2} \right )}{h}$

$f'(x)=\lim\limits_{h\rightarrow 0}\frac{-\sin\left ( \frac{2x+h-\pi/4 }{2}\right )\sin\left ( \frac{h}{2} \right )}{\frac{h}{2}}$

$f'(x)=\sin\left ( \frac{2x+0-\pi/4}{2} \right )\times 1$

$f'(x)=-\sin\left (x-\pi/8 \right )$

Question 2: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): (x+a )

Answer:

Given

f(x)= x + a

As we know, the property,

$f'(x^n)=nx^{n-1}$

Applying that property, we get

$f'(x)=1+0$

$f'(x)=1$

Question 3: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\left(px + q\right) \left ( \frac{r}{x}+ s \right )$

Answer:

Given

$f(x)=\left(px + q\right) \left ( \frac{r}{x}+ s \right )$

$f(x)=pr+psx+\frac{qr}{x}+qs$

As we know, the property,

$f'(x^n)=nx^{n-1}$

Applying that property, we get

$f'(x)=0 + ps +\frac{-qr}{x^2}+0$

$f'(x)=ps + q \left ( \frac{-r}{x^2} \right )$

$f'(x)=ps - \left ( \frac{qr}{x^2} \right )$

Question 4: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $( ax + b ) ( cx + d )^2$

Answer:

Given,

$f(x)=( ax + b ) ( cx + d )^2$

$f(x)=( ax + b ) ( c^2x^2+2cdx+d^2)$

$f(x)=ac^2x^3+2acdx^2+ad^2x+bc^2x^2+2bcdx+bd^2$

Now,

As we know, the property,

$f'(x^n)=nx^{n-1}$

And the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying that property, we get

$f'(x)=3ac^2x^2+4acdx+ad^2+2bc^2x+2bcd+0$

$f'(x)=3ac^2x^2+4acdx+ad^2+2bc^2x+2bcd$

$f'(x)=3ac^2x^2+(4acd+2bc^2)x+ad^2+2bcd$

Question 5: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $ax + b / cx + d$

Answer:

Given,

$f(x)=\frac{ax+b}{cx+d}$

Now, as we know, the derivative of any function

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Hence, the derivative of f(x) is

$\frac{d(\frac{ax+b}{cx+d})}{dx}=\frac{(cx+d)d(\frac{d(ax+b)}{dx})-(ax+b)(\frac{d(cx+d)}{dx})}{(cx+d)^2}$

$\frac{d(\frac{ax+b}{cx+d})}{dx}=\frac{(cx+d)a-(ax+b)c}{(cx+d)^2}$

$\frac{d(\frac{ax+b}{cx+d})}{dx}=\frac{acx+ad-acx-bc}{(cx+d)^2}$

$\frac{d(\frac{ax+b}{cx+d})}{dx}=\frac{ad-bc}{(cx+d)^2}$

Hence, the Derivative of the function is

$\frac{ad-bc}{(cx+d)^2}$ .

Question 6: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):

$\frac{1 + \frac{1}{x}}{1- \frac{1}{x}}$

Answer:

Given,

$f(x)=\frac{1 + \frac{1}{x}}{1- \frac{1}{x}}$

can also be written as

$f(x)=\frac{x+1}{x-1}$

Now, as we know, the derivative of any function

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Hence, the derivative of f(x) is

$\frac{d(\frac{x+1}{x-1})}{dx}=\frac{(x-1)d(\frac{d(x+1)}{dx})-(x+1)(\frac{d(x-1)}{dx})}{(x-1)^2}$

$\frac{d(\frac{x+1}{x-1})}{dx}=\frac{(x-1)1-(x+1)1}{(x-1)^2}$

$\frac{d(\frac{x+1}{x-1})}{dx}=\frac{x-1-x-1}{(x-1)^2}$

$\frac{d(\frac{x+1}{x-1})}{dx}=\frac{-2}{(x-1)^2}$

Hence, the Derivative of the function is

$\frac{-2}{(x-1)^2}$

Question 7: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):

$\frac{1 }{ax ^2 + bx + c}$

Answer:

Given,

$f(x)=\frac{1 }{ax ^2 + bx + c}$

Now, as we know, the derivative of any such function is given by

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Hence, the derivative of f(x) is

$\frac{d(\frac{1}{ax^2+bx+c})}{dx}=\frac{(ax^2+bx+c)d(\frac{d(1)}{dx})-1(\frac{d(ax^2+bx+c)}{dx})}{(ax^2+bx+c)^2}$

$\frac{d(\frac{1}{ax^2+bx+c})}{dx}=\frac{0-(2ax+b)}{(ax^2+bx+c)^2}$

$\frac{d(\frac{1}{ax^2+bx+c})}{dx}=-\frac{(2ax+b)}{(ax^2+bx+c)^2}$

Question 8: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):

$\frac{ax + b }{px^2 + qx + r }$

Answer:

Given,

$f(x)=\frac{ax + b }{px^2 + qx + r }$

Now, as we know, the derivative of any function

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Hence, the derivative of f(x) is

$\frac{d(\frac{ax+b}{px^2+qx+r})}{dx}=\frac{(px^2+qx+r)d(\frac{d(ax+b)}{dx})-(ax+b)(\frac{d(px^2+qx+r)}{dx})}{(px^2+qx+r)^2}$

$\frac{d(\frac{ax+b}{px^2+qx+r})}{dx}=\frac{(px^2+qx+r)a-(ax+b)(2px+q)}{(px^2+qx+r)^2}$

$\frac{d(\frac{ax+b}{px^2+qx+r})}{dx}=\frac{apx^2+aqx+ar-2apx^2-aqx-2bpx-bq}{(px^2+qx+r)^2}$

$\frac{d(\frac{ax+b}{px^2+qx+r})}{dx}=\frac{-apx^2+ar-2bpx-bq}{(px^2+qx+r)^2}$

Question 9: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):

$\frac{px^2 + qx + r }{ax +b }$

Answer:

Given,

$f(x)=\frac{px^2 + qx + r }{ax +b }$

Now, as we know, the derivative of any function

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Hence, the derivative of f(x) is

$\frac{d(\frac{px^2+qx+r}{ax+b})}{dx}=\frac{(ax+b)d(\frac{d(px^2+qx+r)}{dx})-(px^2+qx+r)(\frac{d(ax+b)}{dx})}{(ax+b)^2}$

$\frac{d(\frac{px^2+qx+r}{ax+b})}{dx}=\frac{(ax+b)(2px+q)-(px^2+qx+r)(a)}{(ax+b)^2}$

$\frac{d(\frac{px^2+qx+r}{ax+b})}{dx}=\frac{2apx^2+aqx+2bpx+bq-apx^2-aqx-ar}{(ax+b)^2}$

$\frac{d(\frac{px^2+qx+r}{ax+b})}{dx}=\frac{apx^2+2bpx+bq-ar}{(ax+b)^2}$

Question 10: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):

$\frac{a}{x^4} - \frac{b}{x^2 } + \cos x$

Answer:

Given

$f(x)=\frac{a}{x^4} - \frac{b}{x^2 } + \cos x$

As we know, the property,

$f'(x^n)=nx^{n-1}$

And the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying that property, we get

$f'(x)=\frac{-4a}{x^5} - (\frac{-2b}{x^3 }) +( -\sin x)$

$f'(x)=\frac{-4a}{x^5} +\frac{2b}{x^3 } -\sin x$

Question 11: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers): $4 \sqrt x - 2$

Answer:

Given

$f(x)=4 \sqrt x - 2$

It can also be written as

$f(x)=4 x^{\frac{1}{2}} - 2$

Now,

As we know, the property,

$f'(x^n)=nx^{n-1}$

And the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying that property, we get

$f'(x)=4 (\frac{1}{2})x^{-\frac{1}{2}} - 0$

$f'(x)=2x^{-\frac{1}{2}}$

$f'(x)=\frac{2}{\sqrt{x}}$

Question 12: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):

$( ax + b ) ^ n$

Answer:

Given

$f(x)=( ax + b ) ^ n$

Now, ass we know the chain rule of derivative,

$[f(g(x))]'=f'(g(x))\times g'(x)$

And, the property,

$f'(x^n)=nx^{n-1}$

Also the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying those properties, we get,

$f'(x)=n(ax+b)^{n-1}\times a$

$f'(x)=an(ax+b)^{n-1}$

Question 13: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $( ax + b ) ^ n ( cx + d ) ^ m$

Answer:

Given

$f(x)=( ax + b ) ^ n ( cx + d ) ^ m$

Now, ass we know the chain rule of derivative,

$[f(g(x))]'=f'(g(x))\times g'(x)$

And the Multiplication property of the derivative,

$\frac{d(y_1y_2)}{dx}=y_1\frac{dy_2}{dx}+y_2\frac{dy_1}{dx}$

And, the property,

$f'(x^n)=nx^{n-1}$

Also the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying those properties, we get,

$f'(x)=(ax+b)^n(m(cx+d)^{m-1})+(cx+d)(n(ax+b)^{n-1})$

$f'(x)=m(ax+b)^n(cx+d)^{m-1}+n(cx+d)(ax+b)^{n-1}$

Question 14: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers): $\sin ( x + a )$

Answer:

Given,

$f(x)=\sin ( x + a )$

Now, ass we know the chain rule of derivative,

$[f(g(x))]'=f'(g(x))\times g'(x)$

Applying this property, we get,

$f'(x)=\cos ( x + a )\times 1$

$f'(x)=\cos ( x + a )$

Question 15: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\csc x \cot x$

Answer:

Given,

$f(x)=\csc x \cot x$

The Multiplication Property of the derivative,

$\frac{d(y_1y_2)}{dx}=y_1\frac{dy_2}{dx}+y_2\frac{dy_1}{dx}$

Applying the property

$\frac{d(\csc x)(\cot x))}{dx}=\csc x\frac{d\cot x}{dx}+\cot x\frac{d\csc x}{dx}$

$\frac{d(\csc x)(\cot x))}{dx}=\csc x(-\csc^2x)+\cot x(-\csc x \cot x)$

$\frac{d(\csc x)(\cot x))}{dx}=-\csc^3x-\cot^2 x\csc x$

Hence, the derivative of the function is $-\csc^3x-\cot^2 x\csc x$.

Question 16: Find the derivative of the following functions (it is to be understood that a, b, c, d,p, q, r, and s are fixed non-zero constants and m and n are integers):

$\frac{\cos x }{1+ \sin x }$

Answer:

Given,

$f(x)=\frac{\cos x }{1+ \sin x }$

Now, as we know, the derivative of any function

$\frac{d(\frac{\cos x}{1+\sin x})}{dx}=\frac{(1+\sin x )d(\frac{d\cos x}{dx})-\cos x(\frac{d(1+\sin x)}{dx})}{(1+\sin x)^2}$

Hence, the derivative of f(x) is

$\frac{d(\frac{\cos x}{1+\sin x})}{dx}=\frac{(1+\sin x )(-\sin x)-\cos x(\cos x)}{(1+\sin x)^2}$

$\frac{d(\frac{\cos x}{1+\sin x})}{dx}=\frac{-\sin x-\sin^2 x -\cos^2 x}{(1+\sin x)^2}$

$\frac{d(\frac{\cos x}{1+\sin x})}{dx}=\frac{-\sin x-1}{(1+\sin x)^2}$

$\frac{d(\frac{\cos x}{1+\sin x})}{dx}=-\frac{1}{(1+\sin x)}$

Question 17: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\frac{\sin x + \cos x }{\sin x - \cos x }$

Answer:

Given

$f(x)=\frac{\sin x + \cos x }{\sin x - \cos x }$

can also be written as

$f(x)=\frac{\tan x + 1 }{\tan x - 1 }$

which further can be written as

$f(x)=-\frac{\tan x + tan(\pi/4) }{1-\tan(\pi/4)\tan x }$

$f(x)=-\tan(x-\pi/4)$

Now,

$f'(x)=-\sec^2(x-\pi/4)$

$f'(x)=-\frac{1}{\cos^2(x-\pi/4)}$

Question 18: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):

$\frac{\sec x -1}{\sec x +1}$

Answer:

Given,

$f(x)=\frac{\sec x -1}{\sec x +1}$

which also can be written as

$f(x)=\frac{1-\cos x}{1+\cos x}$

Now,

As we know, the derivative of such a function

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

So, the derivative of the function is,

$\frac{d(\frac{1-\cos x}{1+\cos x})}{dx}=\frac{(1+\cos x)d(\frac{d(1-\cos x)}{dx})-(1-\cos x)(\frac{d(1+\cos x)}{dx})}{(1+\cos x)^2}$

$\frac{d(\frac{1-\cos x}{1+\cos x})}{dx}=\frac{(1+\cos x)(-(-\sin x))-(1-\cos x)(-\sin x)}{(1+\cos x)^2}$

$\frac{d(\frac{1-\cos x}{1+\cos x})}{dx}=\frac{\sin x+\sin x\cos x+\sin x-\cos x\sin x}{(1+\cos x)^2}$

$\frac{d(\frac{1-\cos x}{1+\cos x})}{dx}=\frac{2\sin x}{(1+\cos x)^2}$

This can also be written as

$\frac{d(\frac{1-\cos x}{1+\cos x})}{dx}=\frac{2\sec x\tan x}{(1+\sec x)^2}$ .

Question 19: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\sin^ n x$

Answer:

Given,

$f(x)=\sin^ n x$

Now, as we know the chain rule of derivative,

$[f(g(x))]'=f'(g(x))\times g'(x)$

And, the property,

$f'(x^n)=nx^{n-1}$

Applying those properties, we get

$f'(x)=n\sin^ {n-1} x \cos x$

Hence Derivative of the given function is $n\sin^ {n-1} x \cos x$

Question 20: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):

$\frac{a + b \sin x }{c+ d \cos x }$

Answer:

Given Function

$f(x)=\frac{a + b \sin x }{c+ d \cos x }$

Now, as we know, the derivative of any function of this type is:

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Hence, the derivative of the given function will be:

$\frac{d(\frac{a+b\sin x}{c+d\cos x})}{dx}=\frac{(c+d\cos x)(\frac{d(a+b\sin x)}{dx})-(a+b\sin x)(\frac{d(c+d\cos x x)}{dx})}{(c+d\cos x)^2}$

$\frac{d(\frac{a+b\sin x}{c+d\cos x})}{dx}=\frac{(c+d\cos x)(b\cos x)-(a+b\sin x)(d(-\sin x))}{(c+d\cos x)^2}$

$\frac{d(\frac{a+b\sin x}{c+d\cos x})}{dx}=\frac{cb\cos x+bd\cos^2 x+ad\sin x+bd\sin^2 x}{(c+d\cos x)^2}$

$\frac{d(\frac{a+b\sin x}{c+d\cos x})}{dx}=\frac{cb\cos x+ad\sin x+bd(\sin^2 x+\cos^2 x)}{(c+d\cos x)^2}$

$\frac{d(\frac{a+b\sin x}{c+d\cos x})}{dx}=\frac{cb\cos x+ad\sin x+bd}{(c+d\cos x)^2}$

Question 21: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):

$\frac{\sin ( x+a )}{ \cos x }$

Answer:

Given,

$f(x)=\frac{\sin ( x+a )}{ \cos x }$

Now, as we know, the derivative of any function

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Hence, the derivative of the given function is:

$\frac{d(\frac{\sin(x+a)}{\cos x})}{dx}=\frac{(\cos x)(\frac{d(\sin (x+a))}{dx})-\sin(x+a)(\frac{d(\cos x)}{dx})}{(\cos x)^2}$

$\frac{d(\frac{\sin(x+a)}{\cos x})}{dx}=\frac{(\cos x)(\cos(x+a))-\sin(x+a)(-\sin (x))}{(\cos x)^2}$

$\frac{d(\frac{\sin(x+a)}{\cos x})}{dx}=\frac{(\cos x)(\cos(x+a))+\sin(x+a)(\sin (x))}{(\cos x)^2}$

$\frac{d(\frac{\sin(x+a)}{\cos x})}{dx}=\frac{\cos (x+a-x)}{(\cos x)^2}$

$\frac{d(\frac{\sin(x+a)}{\cos x})}{dx}=\frac{\cos (a)}{(\cos x)^2}$

Question 22: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers): $x ^ 4 ( 5 \sin x - 3 \cos x )$

Answer:

Given

$f(x)=x ^ 4 ( 5 \sin x - 3 \cos x )$

Now, as we know, the Multiplication property of the derivative,

$\frac{d(y_1y_2)}{dx}=y_1\frac{dy_2}{dx}+y_2\frac{dy_1}{dx}$

Hence, the derivative of the given function is:

$\frac{d(x ^ 4 ( 5 \sin x - 3 \cos x ))}{dx}=x^4\frac{d ( 5 \sin x - 3 \cos x )}{dx}+ ( 5 \sin x - 3 \cos x )\frac{dx^4}{dx}$

$\frac{d(x ^ 4 ( 5 \sin x - 3 \cos x ))}{dx}=x^4(5\cos x+3\sin x)+ ( 5 \sin x - 3 \cos x )4x^3$

$\frac{d(x ^ 4 ( 5 \sin x - 3 \cos x ))}{dx}=5x^4\cos x+3x^4\sin x+ 20x^3 \sin x - 12x^3 \cos x$

Question 23: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $( x^2 +1 ) \cos x$

Answer:

Given

$f(x)=( x^2 +1 ) \cos x^{}$

Now, as we know the product rule of derivative,

$\frac{d(y_1y_2)}{dx}=y_1\frac{dy_2}{dx}+y_2\frac{dy_1}{dx}$

The derivative of the given function is

$\frac{d(( x^2 +1 ) \cos x)}{dx}=( x^2 +1 ) \frac{d\cos x}{dx}+\cos x\frac{d( x^2 +1 ) }{dx}$

$\frac{d(( x^2 +1 ) \cos x)}{dx}=( x^2 +1 ) (-\sin x)+\cos x(2x)$

$\frac{d(( x^2 +1 ) \cos x)}{dx}= -x^2 \sin x-\sin x+2x\cos x$

Question 24: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $( ax ^2 + \sin x ) ( p + q \cos x )$

Answer:

Given,

$f(x)=( ax ^2 + \sin x ) ( p + q \cos x )$

Now, as we know, the Multiplication property of the derivative (the product rule)

$\frac{d(y_1y_2)}{dx}=y_1\frac{dy_2}{dx}+y_2\frac{dy_1}{dx}$

And also the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying those properties, we get,

$\frac{d(( ax ^2 + \sin x ) ( p + q \cos x ))}{dx}=(ax^2+\sin x)\frac{d(p+q\cos x)}{dx}+(p+qx)\frac{d(ax^2+sinx)}{dx}$

$\\\frac{d((ax ^2+\sin x ) ( p + q \cos x ))}{dx}=(ax^2+\sin x)(-q\sin x)+(p+qx)(2ax+\cos x)$

$f'(x)=-aqx^2\sin x-q\sin^2 x+2apx+p\cos x+2aqx^2+qx\cos x$

$f'(x)=x^2(-aq\sin x+2aq) +x(2ap+q\cos x)+p\cos x-q\sin^2 x$

Question 25: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $( x+ \cos x ) ( x - \tan x )$

Answer:

Given,

$f(x)=( x+ \cos x ) ( x - \tan x )$

And the Multiplication property of the derivative,

$\frac{d(y_1y_2)}{dx}=y_1\frac{dy_2}{dx}+y_2\frac{dy_1}{dx}$

Also the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying those properties, we get,

$\frac{d(( x+ \cos x ) ( x - \tan x ))}{dx}=(x+\cos x)\frac{d(x-\tan x)}{dx}+(x-\tan x)\frac{d(x+\cos x)}{dx}$

$=(x+\cos x)(1-\sec^2x)+(x-\tan x)(1-\sin x)$

$=(x+\cos x)(-\tan^2x)+(x-\tan x)(1-\sin x)$

$=(-\tan^2x)(x+\cos x)+(x-\tan x)(1-\sin x)$

Question 26: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):

$\frac{4x + 5 \sin x }{3x+ 7 \cos x }$

Answer:

Given,

$f(x)=\frac{4x + 5 \sin x }{3x+ 7 \cos x }$

Now, as we know, the derivative of any function

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Also the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying those properties, we get

$\frac{d(\frac{4x+5\sin x}{3x+7\cos x})}{dx}=\frac{(3x+7\cos x)d(\frac{d(4x+5\sin x)}{dx})-(4x+5\sin x)(\frac{d(3x+7\cos x)}{dx})}{(3x+7\cos x)^2}$

$\frac{d(\frac{4x+5\sin x}{3x+7\cos x})}{dx}=\frac{(3x+7\cos x)(4+5\cos x)-(4x+5\sin x)(3-7\sin x)}{(3x+7\cos x)^2}$

$=\frac{12x+28\cos x+15x\cos x+35\cos^2x-12x-15\sin x+28x\sin x+35\sin^2 x}{(3x+7\cos x)^2}$

$=\frac{12x+28\cos x+15x\cos x-12x-15\sin x+28x\sin x+35(\sin^2 x+\cos^2x)}{(3x+7\cos x)^2}$

$=\frac{28\cos x+15x\cos x-15\sin x+28x\sin x+35}{(3x+7\cos x)^2}$

Question 27: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q,,r, and s are fixed non-zero constants and m and n are integers):

$\frac{x ^2 \cos ( \pi /4 )}{\sin x }$

Answer:

Given,

$f(x)=\frac{x ^2 \cos ( \pi /4 )}{\sin x }$

As we know, the derivative of any function

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Hence, the derivative of the given function is

$\frac{d(\frac{x^2\cos(\pi/4)}{\sin x})}{dx}=\frac{(\sin x)d(\frac{d(x^2\cos(\pi/4))}{dx})-(x^2\cos(\pi/4))(\frac{d\sin x}{dx})}{\sin^2x}$

$\frac{d(\frac{x^2\cos(\pi/4)}{\sin x})}{dx}=\frac{(\sin x)(2x\cos (\pi/4))-(x^2\cos(\pi/4))(\cos x)}{\sin^2x}$

$\frac{d(\frac{x^2\cos(\pi/4)}{\sin x})}{dx}=\frac{2x\sin x\cos (\pi/4)-x^2\cos x\cos(\pi/4)}{\sin^2x}$

Question 28: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q,, r, and s are fixed non-zero constants and m and n are integers):

$\frac{x }{ 1+ \tan x }$

Answer:

Given

$f(x)=\frac{x }{ 1+ \tan x }$

Noas As we know, the derivative of any function

$\frac{d(\frac{x}{1+\tan x})}{dx}=\frac{(1+\tan x)d(\frac{dx}{dx})-x(\frac{d(1+\tan x)}{dx})}{(1+\tan x)^2}$

$\frac{d(\frac{x}{1+\tan x})}{dx}=\frac{(1+\tan x)(1)-x(\sec^2x)}{(1+\tan x)^2}$

$\frac{d(\frac{x}{1+\tan x})}{dx}=\frac{1+\tan x-x\sec^2x}{(1+\tan x)^2}$

Question 29: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $( x + \sec x ) ( x - \tan x )$

Answer:

Given

$f(x)=( x + \sec x ) ( x - \tan x )$

Now, ass we know the Multiplication property of the derivative,

$\frac{d(y_1y_2)}{dx}=y_1\frac{dy_2}{dx}+y_2\frac{dy_1}{dx}$

Also the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying those properties, we get,

The derivative of the given function is,

$\frac{d(( x + \sec x ) ( x - \tan x )}{dx}=(x+\sec x)\frac{d(x-\tan x)}{dx}+(x-\tan x)\frac{d(x+\sec x)}{dx}$

$\frac{d(( x + \sec x ) ( x - \tan x )}{dx}=(x+\sec x)(1-\sec^2x)+(x-\tan x)(1+\sec x\tan x)$

Question 30: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q,,r, and s are fixed non-zero constants and m and n are integers):

$\frac{x }{\sin ^ n x }$

Answer:

Given,

$f(x)=\frac{x }{\sin ^ n x }$

As we know, the derivative of any function

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

As the chain rule of derivatives,

$[f(g(x))]'=f'(g(x))\times g'(x)$

Hence, the derivative of the given function is

$\frac{d(\frac{x}{\sin^nx})}{dx}=\frac{\sin^nxd(\frac{dx}{dx})-x(\frac{d(\sin^nx)}{dx})}{sin^{2n}x}$

$\frac{d(\frac{x}{\sin^nx})}{dx}=\frac{\sin^nx(1)-x(n\sin^{n-1}x\times \cos x)}{sin^{2n}x}$

$\frac{d(\frac{x}{\sin^nx})}{dx}=\frac{\sin^nx-x\cos xn\sin^{n-1}x}{sin^{2n}x}$

$\frac{d(\frac{x}{\sin^nx})}{dx}=\frac{\sin^nx-x\cos xn\sin^{n-1}x}{sin^{2n}x}$

$\frac{d(\frac{x}{\sin^nx})}{dx}=\frac{\sin^{n-1}x(\sin x-nx\cos x)}{sin^{2n}x}$

$\frac{d(\frac{x}{\sin^nx})}{dx}=\frac{(\sin x-nx\cos x)}{sin^{n+1}x}$

Also, read,

• Limits And Derivatives NCERT Exercise 12.1 Solutions
• Limits And Derivatives NCERT Exercise 12.2 Solutions
• Limits And Derivatives Miscellaneous Solutions

Class 11 Maths NCERT Chapter 12: Extra Question

Question:
$\displaystyle\lim _{x \rightarrow 0} \operatorname{cosec} x\left(\sqrt{2 \cos ^2 x+3 \cos x}-\sqrt{\cos ^2 x+\sin x+4}\right)$ is

Solution:
Evaluate the limit:
$
\displaystyle\lim_{x \to 0} \csc x \left( \sqrt{2 \cos^2 x + 3 \cos x} - \sqrt{\cos^2 x + \sin x + 4} \right)
$

Rationalise the expression:
$
\displaystyle\lim_{x \to 0} \frac{\csc x \left( (2 \cos^2 x + 3 \cos x) - (\cos^2 x + \sin x + 4) \right)}{\sqrt{2 \cos^2 x + 3 \cos x} + \sqrt{\cos^2 x + \sin x + 4}}
$

Simplify the numerator:
$
\displaystyle\lim_{x \to 0} \frac{1}{\sin x} \cdot \frac{(\cos^2 x + 3 \cos x - 4) - \sin x}{\sqrt{2 \cos^2 x + 3 \cos x} + \sqrt{\cos^2 x + \sin x + 4}}
$

Rewrite the numerator:
$
\displaystyle\lim_{x \to 0} \frac{(\cos x + 4)(\cos x - 1) - \sin x}{\sin x \left( \sqrt{2 \cos^2 x + 3 \cos x} + \sqrt{\cos^2 x + \sin x + 4} \right)}
$

Using half-angle formulas:
$
\displaystyle\lim_{x \to 0} \frac{-2 \sin^2 \frac{x}{2} (\cos x + 4) - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2} \left( \sqrt{2 \cos^2 x + 3 \cos x} + \sqrt{\cos^2 x + \sin x + 4} \right)}
$

Cancel common terms:

$\displaystyle\lim _{x \rightarrow 0} -\frac{\sin \frac{x}{2}(\cos x+4)+\cos \frac{x}{2}}{\cos \frac{x}{2}\left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)}$

Evaluate the limit by plugging $ x = 0 $:
$
= -\frac{1}{2 \sqrt{5}}
$

Hence, the correct answer is $-\frac{1}{2 \sqrt{5}}$.

Limits and Derivatives Class 11 Chapter 12: Topics

Given below are the topics discussed in the NCERT Solutions for class 11, chapter 12, Limits and Derivatives:

• Introduction
• Intuitive Idea of Derivatives
• Limits
• Limits of Trigonometric Functions
• Derivatives

Limits and Derivatives Class 11 Solutions: Important Formulae

Limits

The limit of a function $f(x)$ as $x$ approaches $a$, denoted as $\lim\limits_{x \rightarrow a} f(x)$, exists when both the left-hand limit and the right-hand limit exist and are equal:

$\lim\limits_{x \rightarrow a} f(x)$ exists if $\lim\limits_{x \rightarrow a^{-}} f(x)=\lim\limits_{x \rightarrow a^{+}} f(x)$

Left-Hand Limit:

$\lim\limits_{x \rightarrow a^{-}} f(x)=\lim\limits_{h \rightarrow 0} f(a-h)$

Right-Hand Limit:

$\lim\limits_{x \rightarrow a^{+}} f(x)=\lim\limits_{h \rightarrow 0} f(a+h)$

Properties of Limits:

If $\lim\limits_{x \rightarrow a} f(x)$ and $\lim\limits_{x \rightarrow a} g(x)$ both exist:
Sum Rule:

$\lim\limits_{x \rightarrow a}[f(x) \pm g(x)]=\lim\limits_{x \rightarrow a} f(x) \pm \lim\limits_{x \rightarrow a} g(x)$

Constant Rule:

$\lim\limits_{x \rightarrow a}[k \cdot f(x)]=k \cdot \lim\limits_{x \rightarrow a} f(x)$

Product Rule:

$\lim\limits_{x \rightarrow a}[f(x) \cdot g(x)]=\lim\limits_{x \rightarrow a} f(x) \cdot \lim\limits _{x \rightarrow a} g(x)$

Quotient Rule:

$\lim\limits_{x \rightarrow a}\left[\frac{f(x)}{g(x)}\right]=\frac{\lim\limits_{x \rightarrow a} f(x)}{\lim\limits_{x \rightarrow a} g(x)}, \quad$ provided $\lim\limits_{x \rightarrow a} g(x) \neq 0$

Standard Limits:

$\lim\limits_{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}$
$\lim\limits_{x \rightarrow 0} \frac{\sin x}{x}=1$
$\lim\limits_{x \rightarrow 0} \frac{\tan x}{x}=1$
$\lim\limits_{x \rightarrow 0} \frac{a^x-1}{x}=\ln a$
$\lim\limits_{x \rightarrow 0} \frac{e^x-1}{x}=1$
$\lim\limits_{x \rightarrow 0} \frac{\ln (1+x)}{x}=1$

Derivatives:

The derivative of a function $f$ at point $x$ is given by:

$f^{\prime}(x)=\lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

Properties of Derivatives:

If $f(x)$ and $g(x)$ are differentiable:
Sum Rule:

$\frac{d}{d x}[f(x)+g(x)]=\frac{d}{d x} f(x)+\frac{d}{d x} g(x)$

Difference Rule:

$\frac{d}{d x}[f(x)-g(x)]=\frac{d}{d x} f(x)-\frac{d}{d x} g(x)$

Product Rule:

$\frac{d}{d x}[f(x) \cdot g(x)]=f^{\prime}(x) g(x)+f(x) g^{\prime}(x)$

Quotient Rule:

$\frac{d}{d x}\left[\frac{f(x)}{g(x)}\right]=\frac{f^{\prime}(x) g(x)-f(x) g^{\prime}(x)}{[g(x)]^2}$

Standard Derivatives:

$\frac{d}{d x}\left(x^n\right)=n x^{n-1}$
$\frac{d}{d x}(\sin x)=\cos x$
$\frac{d}{d x}(\cos x)=-\sin x$
$\frac{d}{d x}(\tan x)=\sec ^2 x$
$\frac{d}{d x}(\cot x)=-\csc ^2 x$
$\frac{d}{d x}(\sec x)=\sec x \cdot \tan x$
$\frac{d}{d x}(\csc x)=-\csc x \cdot \cot x$
$\frac{d}{d x}\left(a^x\right)=a^x \cdot \ln a$
$\frac{d}{d x}\left(e^x\right)=e^x$
$\frac{d}{d x}(\ln x)=\frac{1}{x}$

Approach to Solve Questions of Limits and Derivatives Class 11

Here are some approaches that students can use to answer questions related to limits and derivatives.

• Get familiar with the properties of limits and derivatives, such as the sum rule, constant rule, product rule, quotient rule, etc.
• Get clarity about the standard limits and derivatives formulas.
• Check the indeterminate forms in limits first.
• Before applying the rules of limits or derivatives, simplify the expression or solve it part by part.
• Decide which is better to use for derivatives: the first principle of the standard rules.
• In higher classes, practice using factorisation or L'Hospital's rule to resolve indeterminate forms.
• Prepare a chart of all the standard derivatives in a notebook, which you can revise from time to time for quick recall.

What Extra Should Students Study Beyond NCERT for JEE?

NCERT Solutions for Class 11 Mathematics - Chapter Wise

We at Careers360 compiled all the NCERT class 11 Maths solutions in one place for easy student reference. The following links will allow you to access them

Also Read,

• NCERT Exemplar Class 11 Maths Solutions Chapter 12 Limits and Derivatives
• NCERT Class 11 Maths Chapter 12 Notes Limits and Derivatives

NCERT Solutions for Class 11- Subject Wise

Students can check the following links for more in-depth learning.

• NCERT solutions for class 11 biology
• NCERT solutions for class 11 maths
• NCERT solutions for class 11 chemistry
• NCERT Solutions for Class 11 Physics

NCERT Books and NCERT Syllabus

Here is the latest NCERT syllabus, that is useful for students before strategising their study plan. Also, it contains links to some reference books which are important for further studies.

• NCERT Books Class 11 Maths
• NCERT Syllabus Class 11 Maths
• NCERT Books Class 11
• NCERT Syllabus Class 11