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.

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Limits and Derivatives Class 11 Questions And Answers
Limits and Derivatives Class 11 Questions And Answers PDF Free Download
Limits and Derivatives Class 11 Solutions - Important Formulae
Limits and Derivatives Class 11 NCERT Solutions (Intext Questions and Exercise)
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NCERT Solutions for Class 11 Mathematics - Chapter Wise
Key Features Of Limits And Derivatives Class 11 Solutions
NCERT Solutions for Class 11- Subject Wise
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NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives

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) [(a^x - 1)/x] = ln(a)
lim(x→0) [(e^x - 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)] = a^x * ln(a)
[d/dx (e^x)] = e^x
[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,

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

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,

Let's put

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

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

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

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: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 $\lim_{x \rightarrow a _ 1 }$ f (x) ? For some $a \neq a _ 1 , a _ 2 .... a _n$ , compute l $\lim_{ x \rightarrow a } f (x)$

Answer:

Given,

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

Now,

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

Hence

, $\lim_{x \rightarrow a _ 1 }f(x)=0$

Now,

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

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

Hence

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

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 $\lim_{x \rightarrow a } f (x)$ 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: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 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: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, 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: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, As 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 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: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

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

Also can 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 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

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

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

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

Now, 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, 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

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

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

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

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

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

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

chapter-1	Sets
chapter-2	Relations and Functions
chapter-3	Trigonometric Functions
chapter-4	Mathematical Induction
chapter-5	Complex Numbers and Quadratic equations
chapter-6	Linear Inequalities
chapter-7	Permutation and Combinations
chapter-8	Binomial Theorem
chapter-9	Sequences and Series
chapter-10	Straight Lines
chapter-11	Conic Section
chapter-12	Three Dimensional Geometry
chapter-13	Limits and Derivatives
chapter-14	Mathematical Reasoning
chapter-15	Statistics
chapter-16	Probability

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:

Begin by familiarizing yourself with the basic concepts covered in the chapter.
Understand the fundamental topics of calculus, including both differential and integral calculus.
Memorize the essential formulas that are used in the chapter.
Focus on learning about limits and their properties in detail.
Master the fundamental theorem of calculus and its significance.
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.

Also Read

NCERT Solutions for Exercise 10.1 Class 11 Maths Chapter 10 - Straight Lines

NCERT Solutions for Miscellaneous Exercise Chapter 16 Class 11 - Probability

NCERT Solutions for Miscellaneous Exercise Chapter 13 Class 11 - Limits and Derivatives

NCERT Solutions for Miscellaneous Exercise Chapter 11 Class 11 - Conic Section

NCERT Solutions for Miscellaneous Exercise Chapter 10 Class 11 - Straight Lines

NCERT Solutions for Miscellaneous Exercise Chapter 9 Class 11 - Sequences and Series

NCERT Book for class 11 Biology 2023 - Download PDF

NCERT Books for Class 11 - Download All Subjects PDF here

NCERT Books for class 11 Hindi 2023 - Download PDF Here

NCERT Books for class 11 Chemistry 2023 - Download Pdf here

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NCERT Class 11th Notes Subject wise - Download Free PDF

Environmental Chemistry Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 14 Notes - Download PDF

Hydrocarbons Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 13 Notes - Download PDF

Organic Chemistry- Some Basic Principles and Techniques Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 12 Notes - Download PDF

The P-Block Elements Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 11 Notes - Download PDF

NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 14 Mathematical Reasoning

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

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NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives

Limits and Derivatives Class 11 Questions And Answers

Limits and Derivatives Class 11 Questions And Answers PDF Free Download

Limits and Derivatives Class 11 Solutions - Important Formulae

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

limits and derivatives ncert solutions - Topics

NCERT Solutions for Class 11 Mathematics - Chapter Wise

Key Features Of Limits And Derivatives Class 11 Solutions

NCERT Solutions for Class 11- Subject Wise

NCERT Books and NCERT Syllabus

Frequently Asked Question (FAQs)

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