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    NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives

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    NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives

    Edited By Ramraj Saini | Updated on Sep 25, 2023 07:42 PM IST

    Limits and Derivatives Class 11 Questions And Answers

    NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives are provided here. This Class 11 NCERT syllabus introduces an important area of mathematics called calculus. This is a branch of mathematics which deal with the study of change in the value of a function as the points in the domain change. In this article, you will get limits and derivatives class 11 NCERT solutions that are prepared by highly qualified teachers keeping in mind latest syllabus of CBSE 2023, simple and provide step by step solution to each problem. Students can get all NCERT solutions at one place and practice them to develop indepth understanding of the concepts.

    In class 11 maths chapter 13 question answer, you will get questions related to all the above topics. This chapter is very important for both class 11 final examination and various competitive exams like JEE Main, JEE Advanced, VITEEE, BITSAT, etc. There are 43 questions in 2 exercises limits and derivatives class 11. The first exercise of this chapter deals with problems on limits and the second exercise deals with problems on derivatives. Students can find all NCERT solution for class 11 at one pace for hare and practice them.

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    Limits and Derivatives Class 11 Questions And Answers PDF Free Download

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    Limits and Derivatives Class 11 Solutions - Important Formulae

    Limits:

    The limit of a function f(x) as x approaches a, denoted as lim(x→a) f(x), exists when both the left-hand limit (lim(x→a-) f(x)) and right-hand limit (lim(x→a+) f(x)) exist and are equal (lim(x→a-) f(x) = lim(x→a+) f(x)).

    Left-Hand and Right-Hand Limits:

    Left-Hand Limit: f(a-0) = lim(x→a-) f(x) = lim(h→0) f(a-h)

    Right-Hand Limit: f(a+0) = lim(x→a+) f(x) = lim(h→0) f(a+h)

    Properties of Limits:

    If lim(x→a) f(x) and lim(x→a) g(x) both exist:

    • Sum Rule: lim(x→a) [f(x) ± g(x)] = lim(x→a) f(x) ± lim(x→a) g(x)

    • Constant Rule: lim(x→a) kf(x) = k lim(x→a) f(x)

    • Product Rule: lim(x→a) [f(x) * g(x)] = lim(x→a) f(x) * lim(x→a) g(x)

    • Quotient Rule: lim(x→a) [f(x)/g(x)] = [lim(x→a) f(x)] / [lim(x→a) g(x)] (provided lim(x→a) g(x) ≠ 0)

    Standard Limits:

    • lim(x→a) [(xn - an)/(x - a)] = na(n-1)

    • lim(x→0) [(sin(x))/x] = 1

    • lim(x→0) [(tan(x))/x] = 1

    • lim(x→0) [(ax - 1)/x] = ln(a)

    • lim(x→0) [(ex - 1)/x] = 1

    • lim(x→0) [(ln(1+x))/x] = 1

    Derivatives:

    The derivative of a function f at a point x, denoted as f'(x), is defined as:

    • f'(x) = lim(h→0) [(f(x+h) - f(x))/h]

    Properties of Derivatives:

    If f(x) and g(x) have derivatives in a common domain:c

    • Sum Rule: [d/dx (f(x) + g(x))] = [d/dx (f(x))] + [d/dx (g(x))]

    • Difference Rule: [d/dx (f(x) - g(x))] = [d/dx (f(x))] - [d/dx (g(x))]

    • Product Rule: [d/dx (f(x) * g(x))] = [d/dx f(x)] * g(x) + f(x) * [d/dx g(x)]

    • Quotient Rule: [d/dx (f(x)/g(x))] = ([d/dx f(x)] * g(x) - f(x) * [d/dx g(x)]) / [g(x)]2

    Standard Derivatives:

    • [d/dx (x^n)] = n * x(n-1)

    • [d/dx (sin(x))] = cos(x)

    • [d/dx (cos(x))] = -sin(x)

    • [d/dx (tan(x))] = sec2(x)

    • [d/dx (cot(x))] = -csc2(x)

    • [d/dx (sec(x))] = sec(x) * tan(x)

    • [d/dx (csc(x))] = -csc(x) * cot(x)

    • [d/dx (a^x)] = ax * ln(a)

    • [d/dx (e^x)] = ex

    • [d/dx (ln(x))] = 1/x

    Free download NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives for CBSE Exam.

    Limits and Derivatives Class 11 NCERT Solutions (Intext Questions and Exercise)

    Limits and derivatives class 11 questions and answers - Exercise: 13.1

    Question:1 Evaluate the following limits

    \lim_{x\rightarrow 3} x +3

    Answer:

    \lim_{x\rightarrow 3} x +3

    \Rightarrow \lim_{x\rightarrow 3} 3 +3

    \Rightarrow 6 Answer

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

    Answer:

    Below you can find the solution:

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

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

    Answer:

    The limit

    \lim_{r \rightarrow 1} \pi r^2=\pi(1)^2=\pi

    Answer is \pi

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

    Answer:

    The limit

    \lim_{x \rightarrow {4}} \frac{4x+3 }{x-2}

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

    \Rightarrow \frac{19 }{2} (Answer)

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

    Answer:

    The limit

    \lim_{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} (Answer)

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

    Answer:

    The limit

    \lim_{x \rightarrow 0 }\frac{( x+1)^5 -1}{x }

    Lets 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 have became

    \lim_{y \rightarrow 1 }\frac{ y^5 -1}{y-1 }

    Now As we know the property

    \lim_{x \rightarrow 1 }\frac{ x^5 -a^n}{x-a }=na^{n-1}


    \lim_{y \rightarrow 1 }\frac{ y^5 -1}{y-1 }=5(1)^5=5

    Hence,

    \lim_{x \rightarrow 0 }\frac{( x+1)^5 -1}{x }=5

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

    Answer:

    The limit

    \lim_{x \rightarrow 2} \frac{3 x^2 - x -10 }{x^2 -4}

    \Rightarrow \lim_{x \rightarrow 2} \frac{(x-2)(3x+5) }{(x-2)(x+2)}

    \Rightarrow \lim_{x \rightarrow 2} \frac{(3x+5) }{(x+2)}

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

    \Rightarrow \frac{11 }{4} (Answer)

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

    Answer:

    The limit

    \lim_{x \rightarrow 3} \frac{x ^4 -81}{2x^2 -5x -3}

    At x = 2 both numerator and denominator becomes zero, so lets factorise the function

    \lim_{x \rightarrow 3} \frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)}

    \lim_{x \rightarrow 3} \frac{(x+3)(x^2+9)}{(2x+1)}

    Now we can put the limit directly, so

    \lim_{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_{x \rightarrow 0 } \frac{ax +b}{cx+1}

    Answer:

    The limit,

    \lim_{x \rightarrow 0 } \frac{ax +b}{cx+1}

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

    \Rightarrow b (Answer)

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

    Answer:

    The limit

    \lim_{z\rightarrow 1} \frac{z^{1/3}-1}{z^{1/6}-1}

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

    \lim_{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_{z\rightarrow 1} \frac{(z^{(1/6)}-1)(z^{(1/6)}+1)}{z^{1/6}-1}

    \\=\lim_{z\rightarrow 1}(z^{1/6}+1)\\=(1^{1/6}+1)

    =1+1

    =2 (Answer)

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

    Answer:

    The limit:

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

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

    \lim_{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 (Answer)

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

    Answer:

    \lim_{x\rightarrow -2} \frac{\frac{1}{x}+ \frac{x}{2}}{x+2}

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

    \lim_{x\rightarrow -2} \frac{\frac{1}{x}+ \frac{x}{2}}{x+2}

    =\lim_{x\rightarrow -2} \frac{\frac{x+2}{2x}}{x+2}

    =\lim_{x\rightarrow -2} \frac{1}{2x}

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

    = -\frac{1}{4} (Answer)

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

    Answer:

    The limit

    \lim_{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_{x \rightarrow 0 } \frac{\sin ax }{bx }

    =\lim_{x \rightarrow 0 } \frac{\sin ax(ax) }{bx(ax) }

    =\lim_{x \rightarrow 0 } \frac{\sin ax }{ax }\frac{a}{b}

    As \lim_{x\rightarrow 0}\frac{sinx}{x}=1

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

    =\frac{a}{b} (Answer)

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

    Answer:

    The limit,

    \lim_{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_{x\rightarrow {0}}\frac{\frac{sinax}{ax}}{\frac{sinbx}{bx}}\cdot\frac{ax}{bx}

    =\frac{\lim_{ax\rightarrow {0}}\frac{sinax}{ax}}{\lim_{bx\rightarrow {0}}\frac{sinbx}{bx}}\cdot\frac{a}{b}

    =\frac{a}{b} (Answer)

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

    Answer:

    The limit

    \lim_{x \rightarrow \pi } \frac{\sin ( \pi -x )}{\pi ( \pi -x)}

    \Rightarrow \lim_{x \rightarrow \pi } \frac{\sin ( \pi -x )}{ ( \pi -x)}\times\frac{1}{\pi}

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

    = \frac{1}{\pi} (Answer)

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

    Answer:

    The limit

    \lim_{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_{x\rightarrow 0}\frac{\cos x }{\pi -x }

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

    =\frac{1 }{\pi } (Answer)

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

    Answer:

    The limit:

    \lim_{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_{x\rightarrow 0} \frac{\cos 2x -1}{\cos x -1}

    \lim_{x\rightarrow 0} \frac{-2(sin^2x)}{-2(sin^2(\frac{x}{2}))}

    =\lim_{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 (Answer)

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

    Answer:

    \lim_{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 function consist of 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_{x \rightarrow 0}\frac{ax + x \cos x }{b \sin x }

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

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

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

    =\frac{a+1}{b} (Answer)

    Question:19 Evaluate the following limits \lim_{x \rightarrow 0} x \sec x

    Answer:

    \lim_{x \rightarrow 0} x \sec x

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

    \lim_{x \rightarrow 0} x \sec x

    =(0)\times 1

    =0 . (Answer)

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

    Answer:

    \lim_{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_{x\rightarrow 0} \frac{\sin ax + bx }{ax + \sin bx } a,b ,a + b \neq 0

    =\lim_{x\rightarrow 0} \frac{\frac{\sin ax}{ax} \cdot ax+ bx }{ax + \frac{\sin bx}{bx}\cdot bx }

    =\lim_{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 (Answer)

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

    Answer:

    \lim_{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_{x \rightarrow 0} \left ( \csc x - \cot x \right )

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

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

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

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

    =\lim_{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 (Answer)

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

    Answer:

    \lim_{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, limit will also change.

    as

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

    So function in new variable becomes,

    \lim_{y \rightarrow 0 } \frac{\tan 2(y+\frac{\pi}{2}) }{y+\pi/2- \pi /2 }

    =\lim_{y \rightarrow 0 } \frac{\tan (2y+\pi) }{y }

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

    =\lim_{y \rightarrow 0 } \frac{\tan (2y) }{y }

    =\lim_{y \rightarrow 0 } \frac{sin2y}{2y}\cdot\frac{2}{cos2y}

    =1\times 2

    =2 (Answer)

    Question:23 Find \lim_{x \rightarrow 0} f (x ) \: \: \lim_{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_{x\rightarrow{0^-}}f(x)=\lim_{x\rightarrow{0^-}}(2x+3)=2(0)+3=3

    at\:x=0^+

    \lim_{x\rightarrow{0^+}}f(x)=\lim_{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_{x\rightarrow{1^+}}f(x)=\lim_{x\rightarrow{1^+}}3(x+1)=3(1+1)=6

    at\:x=1^-

    \lim_{x\rightarrow{1^-}}f(x)=\lim_{x\rightarrow{1^-}}3(x+1)=3(1+1)=6

    Hence limit at x = 1 is 6.

    Question:24 Find \lim_{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_{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_{x \rightarrow 1^+} f (x ) = \lim_{x \rightarrow 1} (-x^2-1)=-(1)^2-1=-2

    Limit at x=1^-

    \lim_{x \rightarrow 1^-} f (x ) = \lim_{x \rightarrow 1} (x^2-1)=(1)^2-1=0

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

    Question:25 Evaluate \lim_{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_{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_{x \rightarrow 0^+} f (x) = \lim_{x \rightarrow 0^+} \frac{|x|}{x}=\frac{x}{x}=1

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

    \lim_{x \rightarrow 0^-} f (x) = \lim_{x \rightarrow 0^-} \frac{|x|}{x}=\frac{-x}{x}=-1

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

    Question:26 Evaluate \lim_{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_{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_{x \rightarrow 0^+} f (x) = \lim_{x \rightarrow 0^+} \frac{x}{|x|}=\frac{x}{x}=1

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

    \lim_{x \rightarrow 0^-} f (x) = \lim_{x \rightarrow 0^-} \frac{x}{|x|}=\frac{x}{-x}=-1

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

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

    Answer:

    \lim_{x \rightarrow 5 } f (x) , where f( x) = |x| -5

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

    \lim_{x \rightarrow 5^+} f (x) = \lim_{x \rightarrow 5^+} |x|-5=5-5=0

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

    \lim_{x \rightarrow 5^-} f (x) = \lim_{x \rightarrow 5^-}|x|-5=5-5=0

    Since 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_{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_{x\rightarrow 1^-} f(x)= \lim_{x\rightarrow 1}(a+bx)=a+b(1)=a+b=4

    Right-hand limit

    \lim_{x\rightarrow 1^+} f(x)= \lim_{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 value of a and b are 0 and 4 respectively.

    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_{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.

    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_{x \rightarrow 0^+} f (x) = \lim_{x \rightarrow 0^+} |x|-1=1-1=0

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

    \lim_{x \rightarrow 0^-} f (x) = \lim_{x \rightarrow 0^-} |x|+1=0+1=1

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

    Case 2: When a < 0

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

    \lim_{x \rightarrow a^+} f (x) = \lim_{x \rightarrow a^+} |x|-1=a-1

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

    \lim_{x \rightarrow a^-} f (x) = \lim_{x \rightarrow a^-} |x|-1=a-1

    Since LHL = RHL, 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_{x \rightarrow a^+} f (x) = \lim_{x \rightarrow a^+} |x|+1=a+1

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

    \lim_{x \rightarrow a^-} f (x) = \lim_{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_{x \rightarrow 1} \frac{f (x)-2}{x^2-1} = \pi , evaluate \lim_{x \rightarrow 1} f (x)

    Answer:

    Given

    \lim_{x \rightarrow 1} \frac{f (x)-2}{x^2-1} = \pi

    Now,

    \lim_{x \rightarrow 1} \frac{f (x)-2}{x^2-1} = \frac{\lim_{x \rightarrow 1}(f (x)-2)}{\lim_{x \rightarrow 1}(x^2-1)}=\pi

    {\lim_{x \rightarrow 1}(f (x)-2)}=\pi{\lim_{x \rightarrow 1}(x^2-1)}

    {\lim_{x \rightarrow 1}(f (x)-2)}=\pi{(1-1)}

    {\lim_{x \rightarrow 1}(f (x)-2)}=0

    {\lim_{x \rightarrow 1}f (x)}=2

    Question:32 If

    \\f(x)=\begin{bmatrix} mx^2+n\ \ ;x<0 & \\nx+m\ \ ;x \leq0\leq 1t &\\ nx^3+m\ \ ;x>1 \end{bmatrix}

    Answer:

    Given,

    \\f(x)=\begin{bmatrix} mx^2+n\ \ ;x<0 & \\nx+m\ \ ;x \leq0\leq 1t &\\ nx^3+m\ \ ;x>1 \end{bmatrix}

    Case 1: Limit at x = 0

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

    \lim_{x \rightarrow 0^+} f (x) = \lim_{x \rightarrow 0^+} nx+m=n(0)+m=m

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

    \lim_{x \rightarrow 0^-} f (x) = \lim_{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_{x \rightarrow 1^+} f (x) = \lim_{x \rightarrow 1^+} nx^3+m=n(1)^3+m=n+m

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

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

    Hence Limit at 1 exists at all integers.

    Class 11 maths chapter 13 NCERT solutions - Exercise: 13.2

    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_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

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

    f'(10)=\lim_{h\rightarrow 0}\frac{f(10+h)-f(10)}{h}

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

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

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

    f'(10)=\lim_{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_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

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

    f'(1)=\lim_{h\rightarrow 0}\frac{f(1+h)-f(1)}{h}

    f'(1)=\lim_{h\rightarrow 0}\frac{(1+h)-(1)}{h}

    f'(1)=\lim_{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_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

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

    f'(100)=\lim_{h\rightarrow 0}\frac{f(100+h)-f(100)}{h}

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

    f'(100)=\lim_{h\rightarrow 0}\frac{99h}{h}

    f'(100)=99

    Question:4 (i) Find the derivative of the following functions from 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_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}


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

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

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

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

    f'(x)=3x^2

    Question:4.(ii) Find the derivative of the following function from 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_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

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

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

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

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

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

    Question:4(iii) Find the derivative of the following functions from 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_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

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

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

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

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

    f'(x)=\lim_{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 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_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

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

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

    f'(x)=\lim_{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_{h\rightarrow 0}\frac{-2h}{(x-1)(x+h-1)h}

    f'(x)=\lim_{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 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}

    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 first principle.

    Answer:

    Given,

    f(x)= \cos x

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

    f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

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

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

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

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

    f'(x)=\lim_{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 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

    Limits and derivatives NCERT solutions - Miscellaneous Exercise

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

    Answer:

    Given.

    f(x)=-x

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

    f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

    f'(x)=\lim_{h\rightarrow 0}\frac{-(x+h)-(-x)}{h}

    f'(x)=\lim_{h\rightarrow 0}\frac{-h}{h}

    f'(x)=-1

    Question:1(ii) Find the derivative of the following functions from 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_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

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

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

    f'(x)=\lim_{h\rightarrow 0}\frac{\frac{-x+x+h}{x(x+h)}}{h}

    f'(x)=\lim_{h\rightarrow 0}\frac{\frac{h}{(x+h)(x)}}{h}

    f'(x)=\lim_{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 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_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

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

    f'(x)=\lim_{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_{h\rightarrow 0}\frac{2\cos(\frac{2x+h+2}{2})\sin(\frac{h}{2})}{h}

    f'(x)=\lim_{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 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_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

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

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

    f'(x)=\lim_{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: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 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}}

    Also can 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 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: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, 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}

    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, As we know the chain rule of derivative,

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

    And the Multiplication property of 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: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 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 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: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 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}

    Which 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 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 derivative,

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

    Hence 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 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 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 }

    Now, As we know the derivative of any function

    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{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: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, As we know the Multiplication property of 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 }

    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 chain rule of derivative,

    [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}

    limits and derivatives ncert solutions - Topics

    13.1 Introduction

    13.2 Intuitive Idea of Derivatives

    13.3 Limits

    13.4 Limits of Trigonometric Functions

    13.5 Derivatives

    Interested students can practice class 11 maths ch 13 question answer using following exercises.

    Limit of a function(f) at a point is the common value of the left-hand limits and right-hand limits if they coincide.

    For a function 'f' and a real number 'a'. \lim_{x\rightarrow a}\:f(x) and f(a) may not be the same (One may be defined and not the other one).

    For functions 'f' and 'g' the following properties holds:

    \\\lim_{x\rightarrow a} [f(x)\pm g(x)]=\lim_{x\rightarrow a}f(x)\pm \lim_{x\rightarrow a}g(x)\\\:\:\:\\lim_{x\rightarrow a} [f(x). g(x)]=\lim_{x\rightarrow a}f(x). \lim_{x\rightarrow a}g(x)\\\:\:\:\\lim_{x\rightarrow a}[\frac{f(x)}{g(x)}]=\frac{\lim_{x\rightarrow a}f(x)}{\lim_{x\rightarrow a}g(x)}

    • Some of the standard limits are-

    \\\lim_{x\rightarrow a}\frac{x^n-a^n}{x-a}=na^{n-1}\\\:\\lim_{x\rightarrow a}\frac{sin\: x}{x}=1\\\:\\lim_{x\rightarrow a}\frac{1-cos\: x}{x}=0

    • The derivative of a function 'f' at 'a' is defined by-

    f^{'}=\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}

    • The derivative of a function 'f' at any point 'x' is defined by-

    f^{'}(x)=\frac{df(x)}{dx}=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

    NCERT Solutions for Class 11 Mathematics - Chapter Wise

    Key Features Of Limits And Derivatives Class 11 Solutions

    Comprehensive coverage of concepts: The ch 13 maths class 11 provides a thorough and comprehensive coverage of the fundamental concepts of limits and derivatives, which are essential for understanding the principles of calculus.

    Simple and easy language: The class 11 maths limits and derivatives is written in a simple and easy-to-understand language, making it accessible to students of all levels of mathematical aptitude.

    Step-by-step explanations: The chapter provides step-by-step explanations of the concepts, with worked-out examples and exercises that help students to master the topics covered in the chapter.

    NCERT Solutions for Class 11- Subject Wise

    Tip- Limits class 11 is more on the theory-based so you should understand every theorem and try to solve questions based on the theorem. Derivatives required more practice of problems. There are 30 questions in the miscellaneous exercise. You should solve miscellaneous exercises also to get command on this chapter with the help of NCERT solutions for class 11 maths chapter 13 limits and derivatives.

    NCERT Books and NCERT Syllabus

    Frequently Asked Question (FAQs)

    1. What is the effective way to study limits and derivatives class 11 NCERT solutions?

    Effective way to study chapter 13 class 11 maths includes following points:

    1. Begin by familiarizing yourself with the basic concepts covered in the chapter.
    2. Understand the fundamental topics of calculus, including both differential and integral calculus.
    3. Memorize the essential formulas that are used in the chapter.
    4. Focus on learning about limits and their properties in detail.
    5. Master the fundamental theorem of calculus and its significance.
    6. To solidify your understanding of the concepts, practice problems on a regular basis.
    2. What are the topics included in limits and derivatives class 11 solutions?

    class11 maths ch13 includes following topics:

    1. Introduction
    2. Intuitive Idea of Derivatives
    3. Limits
    4. Limits of Trigonometric Functions
    5. Derivatives

    3. How does the NCERT solutions for class 11 chapter 13 maths are helpful ?

    NCERT solutions for class 11 math ch 13 are helpful to the students in solving the NCERT problems as well building foundation for class 12th chapters such as differentiation, integration and many. These solutions are provided in a detailed manner which can be understood by an average student also.

    4. Where can I find the complete class 11 maths NCERT solutions chapter 13?

    Students can get the detailed NCERT solutions for class 11 maths here. they can practice these solutions to build concepts that lead confidence during exam and ultimately help to score well in the exam.

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    Get answers from students and experts

    A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

    Option 1)

    0.34\; J

    Option 2)

    0.16\; J

    Option 3)

    1.00\; J

    Option 4)

    0.67\; J

    A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

    Option 1)

    2.45×10−3 kg

    Option 2)

     6.45×10−3 kg

    Option 3)

     9.89×10−3 kg

    Option 4)

    12.89×10−3 kg

     

    An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

    Option 1)

    2,000 \; J - 5,000\; J

    Option 2)

    200 \, \, J - 500 \, \, J

    Option 3)

    2\times 10^{5}J-3\times 10^{5}J

    Option 4)

    20,000 \, \, J - 50,000 \, \, J

    A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

    Option 1)

    K/2\,

    Option 2)

    \; K\;

    Option 3)

    zero\;

    Option 4)

    K/4

    In the reaction,

    2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

    Option 1)

    11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

    Option 2)

    6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

    Option 3)

    33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

    Option 4)

    67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

    How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

    Option 1)

    0.02

    Option 2)

    3.125 × 10-2

    Option 3)

    1.25 × 10-2

    Option 4)

    2.5 × 10-2

    If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

    Option 1)

    decrease twice

    Option 2)

    increase two fold

    Option 3)

    remain unchanged

    Option 4)

    be a function of the molecular mass of the substance.

    With increase of temperature, which of these changes?

    Option 1)

    Molality

    Option 2)

    Weight fraction of solute

    Option 3)

    Fraction of solute present in water

    Option 4)

    Mole fraction.

    Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

    Option 1)

    twice that in 60 g carbon

    Option 2)

    6.023 × 1022

    Option 3)

    half that in 8 g He

    Option 4)

    558.5 × 6.023 × 1023

    A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

    Option 1)

    less than 3

    Option 2)

    more than 3 but less than 6

    Option 3)

    more than 6 but less than 9

    Option 4)

    more than 9

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    QA Manager

    Quality Assurance Manager Job Description: A QA Manager is an administrative professional responsible for overseeing the activity of the QA department and staff. It involves developing, implementing and maintaining a system that is qualified and reliable for testing to meet specifications of products of organisations as well as development processes. 

    2 Jobs Available
    QA Lead

    A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

    2 Jobs Available
    Reliability Engineer

    Are you searching for a Reliability Engineer job description? A Reliability Engineer is responsible for ensuring long lasting and high quality products. He or she ensures that materials, manufacturing equipment, components and processes are error free. A Reliability Engineer role comes with the responsibility of minimising risks and effectiveness of processes and equipment. 

    2 Jobs Available
    Safety Manager

    A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

    2 Jobs Available
    Corporate Executive

    Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

    2 Jobs Available
    ITSM Manager

    ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks. 

    3 Jobs Available
    Information Security Manager

    Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

    3 Jobs Available
    Computer Programmer

    Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

    3 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    IT Consultant

    An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.

    2 Jobs Available
    Data Architect

    A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements. 

    2 Jobs Available
    Security Engineer

    The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.

    2 Jobs Available
    UX Architect

    A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy. 

     

    2 Jobs Available
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