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

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

Edited By Vishal kumar | Updated on Nov 20, 2023 12:30 PM IST

## 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.

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.

## Access Limits And Derivatives Class 11 Chapter 13-Miscellaneous Exercise

Given.

f(x)=-x

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

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

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

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

$f'(x)=-1$

Given.

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

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

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

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

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

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

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

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

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

Given.

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

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

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

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

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

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

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

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

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

Given.

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

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

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

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

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

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

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

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

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

Given

f(x)= x + a

As we know, the property,

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

applying that property we get

$f'(x)=1+0$

$f'(x)=1$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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,

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

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

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,

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

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

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,

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

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

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 $-\csc^3x-\cot^2 x\csc x$ .

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

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

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

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

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

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,

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

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

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

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

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 $n\sin^ {n-1} x \cos x$

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:

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

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

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

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

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

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:

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

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

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

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

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

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:

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

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

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

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

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

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

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

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,

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

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

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

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

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,

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

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

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

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

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

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

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

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

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

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

Given,

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

Now, As we know the derivative of any function

Now, As we know the derivative of any function

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

Hence the derivative of the given function is

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

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

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

Given

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

Now, As we know the derivative of any function

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

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

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

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,

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

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

Given,

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

Now, As we know the derivative of any function

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

Also chain rule of derivative,

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

Hence the derivative of the given function is

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

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

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

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

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

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

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

Questions related to the derivatives trigonometric functions and polynomial functions are described through 30 questions in the NCERT book Class 11 Maths chapter 13 miscellaneous solutions. All the questions are important and can expect a good number of questions of a similar type for class exams and board exams. Also similar types of questions as in NCERT solutions for Class 11 Maths chapter 13 miscellaneous exercise can be expected for competitive exams like JEE Main and various state-level engineering entrance exams.

Also Read| Limits And Derivatives Class 11 Notes

## 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:

1. Intuitive Idea of Derivatives
2. Limits
3. Limits of Trigonometric Functions
4. Derivatives
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## Key Features of Class 11 Maths ch 13 Miscellaneous Exercise Solutions

1. Comprehensive Solutions: Solutions provided by Careers360 for the miscellaneous exercise class 11 chapter 13 are comprehensive, covering a wide range of topics related to Limits and Derivatives.

2. Expert-Crafted Solutions: Class 11 maths miscellaneous exercise chapter 13 solution meticulously crafted by subject matter experts, ensuring accuracy, clarity, and alignment with the prescribed syllabus. This helps students grasp complex concepts effectively.

3. Free Access: The class 11 chapter 13 miscellaneous exercise solutions are freely accessible to all students, promoting inclusivity and making quality education available without financial constraints.

4. User-Friendly Format: Class 11 maths ch 13 miscellaneous exercise solutions Presented in an easy-to-understand format, featuring step-by-step explanations, diagrams, and relevant formulas. This user-friendly approach aids in better comprehension and retention.

5. Additional Resources: Careers360 offers supplementary study materials and resources, providing valuable aids to reinforce learning and practice in addition to the miscellaneous exercise solutions.

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## Subject Wise NCERT Exampler Solutions

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
1. Function x: derivative=1

2. Function sinx: derivative=cosx

3. 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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