NCERT Solutions for Miscellaneous Exercise Chapter 13 Class 11

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

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NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Miscellaneous Exercise - Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 13 miscellaneous exercise covers the concepts of derivatives and limits. To find the derivative from the first principle the concept of limit is required. All the questions in the miscellaneous example and miscellaneous exercise chapter 13 Class 11 are to find the derivative of a function. The Class 11 Maths Chapter 13 miscellaneous exercise solutions covers the derivative of polynomial functions and also the questions in the Class 11 Maths chapter 13 miscellaneous solutions covers the derivative of trigonometric functions.

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The class 11 chapter 13 maths miscellaneous solutions are meticulously crafted by subject experts, providing in-depth explanations and step-by-step guidance. Additionally, the solutions are available in a downloadable PDF format, offering students the option for offline use without any charge. This class 11 maths ch 13 miscellaneous exercise solutions is designed to facilitate a thorough understanding of conic sections and support convenient and cost-free self-study. Miscellaneous exercise Chapter 13 Class 11 and the following exercises are present in the NCERT Class 11 chapter limits and derivatives.

** In the CBSE Syllabus for 2023-24, this miscellaneous exercise class 11 chapter 13 has been renumbered as Chapter 12.

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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

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

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

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

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

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

NCERT Solutions for Class 11 Maths Chapter 13 Miscellaneous Exercise- Derivatives of Some Important Functions

Following are some important derivatives that help to solve miscellaneous exercise chapter 13 Class 11

Function	Derivative
sinx	cosx
cosx	-sinx
tanx	sec2x
cotx	-cosec2x
secx	Secx tanx
cosecx	-cosecx cotx
xsinx	xcosx+sinx
xn	nxn-1

Topic Covered in class 11 Chapter 13 Maths Miscellaneous Solutions

The class 11 maths miscellaneous exercise chapter 13 is centred around the following topics:

Intuitive Idea of Derivatives
Limits
Limits of Trigonometric Functions
Derivatives

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Key Features of Class 11 Maths ch 13 Miscellaneous Exercise Solutions

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Frequently Asked Questions (FAQs)

1. Write cotx in terms of cos and sin functions

cotx=cosx/sinx

2. Write the derivative of g(x)=cotx

Derivative of g(x) is -cosec^2x

3. Expand sin2z

sin2z=2sinzcosz

4. Give the derivative of sin2x

The required derivative=2(cos^2x-sin^2x)

5. List the derivative of x, sinx, xsinx

Function x: derivative=1
Function sinx: derivative=cosx
Function xsinx; derivative=xcosx+sinx

6. What is the rule used in finding the derivative of x sinx?

The product rule. If f(x)=u and g(x)=v, then (uv)’=uv’+u’v

7. What is [u+v]’ {u =f(x) and v=g(x)

[u+v]’=u’+v’

8. What is the number of problems given in miscellaneous exercise chapter 13 Class 11?

30 questions are given in the NCERT solutions for Class 11 Maths chapter 13 miscellaneous exercise

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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

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

NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Miscellaneous Exercise - Download Free PDF

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More About NCERT Solutions for Class 11 Maths Chapter 13 Miscellaneous Exercise:

NCERT Solutions for Class 11 Maths Chapter 13 Miscellaneous Exercise- Derivatives of Some Important Functions

Topic Covered in class 11 Chapter 13 Maths Miscellaneous Solutions

Key Features of Class 11 Maths ch 13 Miscellaneous Exercise Solutions

NCERT Solutions of Class 11 Subject Wise

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