Class 11 Maths NCERT Solutions Chapter 12 Miscellaneous Exercise – Limits and Derivatives

By definition, f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}.

f(x+h)-f(x)=-(x+h)-(-x)=-h

f'(x)=\lim_{h\to0}\dfrac{-h}{h}=\lim_{h\to0}(-1)=-1

Write f(x)=(-x)^{-1}=-\dfrac{1}{x}.

f(x+h)-f(x)=-\dfrac{1}{x+h}+\dfrac{1}{x}=\dfrac{-x+(x+h)}{x(x+h)}=\dfrac{h}{x(x+h)}

f'(x)=\lim_{h\to0}\dfrac{1}{x(x+h)}=\dfrac{1}{x^2}

f(x+h)-f(x)=\sin(x+h+1)-\sin(x+1)

Using \sin C-\sin D=2\cos\left(\dfrac{C+D}{2}\right)\sin\left(\dfrac{C-D}{2}\right) with C=x+h+1, D=x+1:

=2\cos\left(x+1+\dfrac{h}{2}\right)\sin\left(\dfrac{h}{2}\right)

f'(x)=\lim_{h\to0}\left[\cos\left(x+1+\dfrac{h}{2}\right)\times\dfrac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right]=\cos(x+1)\times1=\cos(x+1)

f(x+h)-f(x)=\cos\left(x+h-\dfrac{\pi}{8}\right)-\cos\left(x-\dfrac{\pi}{8}\right)

Using \cos C-\cos D=-2\sin\left(\dfrac{C+D}{2}\right)\sin\left(\dfrac{C-D}{2}\right) with C=x+h-\frac{\pi}{8}, D=x-\frac{\pi}{8}:

=-2\sin\left(x-\dfrac{\pi}{8}+\dfrac{h}{2}\right)\sin\left(\dfrac{h}{2}\right)

f'(x)=\lim_{h\to0}\left[-\sin\left(x-\dfrac{\pi}{8}+\dfrac{h}{2}\right)\times\dfrac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right]=-\sin\left(x-\dfrac{\pi}{8}\right)

Since a is a constant, its derivative is 0, and the derivative of x is 1:

\dfrac{d}{dx}(x+a)=1+0=1

Expanding first is simplest here:

(px+q)\left(\dfrac{r}{x}+s\right)=pr+psx+\dfrac{qr}{x}+qs

Differentiating term by term:

\dfrac{d}{dx}\left[pr+psx+\dfrac{qr}{x}+qs\right]=ps-\dfrac{qr}{x^2}

Using the product rule with u=ax+b (u'=a) and v=(cx+d)^2 (using the chain rule, v'=2c(cx+d)):

f'(x)=a(cx+d)^2+(ax+b)\cdot2c(cx+d)

Taking out the common factor (cx+d):

f'(x)=(cx+d)\left[a(cx+d)+2c(ax+b)\right]=(cx+d)(3acx+ad+2bc)

Using the quotient rule with u=ax+b (u'=a) and v=cx+d (v'=c):

f'(x)=\dfrac{a(cx+d)-c(ax+b)}{(cx+d)^2}=\dfrac{acx+ad-acx-bc}{(cx+d)^2}

Multiplying numerator and denominator by x first simplifies the function:

\dfrac{1+\frac{1}{x}}{1-\frac{1}{x}}=\dfrac{x+1}{x-1}

Using the quotient rule with u=x+1 (u'=1) and v=x-1 (v'=1):

f'(x)=\dfrac{1(x-1)-(x+1)(1)}{(x-1)^2}=\dfrac{-2}{(x-1)^2}

Using the quotient rule with u=1 (u'=0) and v=ax^2+bx+c (v'=2ax+b):

f'(x)=\dfrac{0(ax^2+bx+c)-1(2ax+b)}{(ax^2+bx+c)^2}

Using the quotient rule with u=ax+b (u'=a) and v=px^2+qx+r (v'=2px+q):

f'(x)=\dfrac{a(px^2+qx+r)-(ax+b)(2px+q)}{(px^2+qx+r)^2}

Expanding the numerator:

apx^2+aqx+ar-\left[2apx^2+aqx+2bpx+bq\right]=-apx^2-2bpx+(ar-bq)

Using the quotient rule with u=px^2+qx+r (u'=2px+q) and v=ax+b (v'=a):

f'(x)=\dfrac{(2px+q)(ax+b)-(px^2+qx+r)(a)}{(ax+b)^2}

Expanding the numerator:

\left[2apx^2+2bpx+aqx+bq\right]-\left[apx^2+aqx+ar\right]=apx^2+2bpx+(bq-ar)

Rewrite as ax^{-4}-bx^{-2}+\cos x and differentiate term by term:

\dfrac{d}{dx}\left[ax^{-4}-bx^{-2}+\cos x\right]=-4ax^{-5}+2bx^{-3}-\sin x

Write 4\sqrt{x}-2=4x^{1/2}-2 and use \dfrac{d}{dx}(x^n)=nx^{n-1} with n=\dfrac{1}{2}:

\dfrac{d}{dx}\left[4x^{1/2}-2\right]=4\times\dfrac{1}{2}x^{-1/2}-0=2x^{-1/2}=\dfrac{2}{\sqrt{x}}

Using the power rule together with the chain rule (the derivative of the "inner" function ax+b is a):

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

Using the product rule with u=(ax+b)^n (u'=na(ax+b)^{n-1}, from Question 12) and v=(cx+d)^m (v'=mc(cx+d)^{m-1}, by the same reasoning):

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

Taking out the common factor (ax+b)^{n-1}(cx+d)^{m-1}:

f'(x)=(ax+b)^{n-1}(cx+d)^{m-1}\left[na(cx+d)+mc(ax+b)\right]

Treating a as a constant, the inner function x+a has derivative 1, so by the chain rule:

\dfrac{d}{dx}\sin(x+a)=\cos(x+a)\times1=\cos(x+a)

Using the product rule with u=\operatorname{cosec}x (u'=-\operatorname{cosec}x\cot x) and v=\cot x (v'=-\operatorname{cosec}^2x):

f'(x)=-\operatorname{cosec}x\cot x\cdot\cot x+\operatorname{cosec}x\cdot\left(-\operatorname{cosec}^2x\right)

=-\operatorname{cosec}x\cot^2x-\operatorname{cosec}^3x=-\operatorname{cosec}x\left(\cot^2x+\operatorname{cosec}^2x\right)

Using the quotient rule with u=\cos x (u'=-\sin x) and v=1+\sin x (v'=\cos x):

f'(x)=\dfrac{-\sin x(1+\sin x)-\cos x\cdot\cos x}{(1+\sin x)^2}=\dfrac{-\sin x-\sin^2x-\cos^2x}{(1+\sin x)^2}

Using \sin^2x+\cos^2x=1:

=\dfrac{-\sin x-1}{(1+\sin x)^2}=\dfrac{-(1+\sin x)}{(1+\sin x)^2}=-\dfrac{1}{1+\sin x}

Using the quotient rule with u=\sin x+\cos x (u'=\cos x-\sin x) and v=\sin x-\cos x (v'=\cos x+\sin x):

f'(x)=\dfrac{(\cos x-\sin x)(\sin x-\cos x)-(\sin x+\cos x)(\cos x+\sin x)}{(\sin x-\cos x)^2}

The first term is -(\cos x-\sin x)^2, and the second is (\sin x+\cos x)^2. Expanding both using 2\sin x\cos x=\sin2x:

-(\cos x-\sin x)^2=-(1-\sin2x) and (\sin x+\cos x)^2=1+\sin2x

So the numerator is -(1-\sin2x)-(1+\sin2x)=-2, and the denominator is (\sin x-\cos x)^2=1-\sin2x:

f'(x)=\dfrac{-2}{1-\sin2x}=\dfrac{2}{\sin2x-1}

Using the quotient rule with u=\sec x-1 (u'=\sec x\tan x) and v=\sec x+1 (v'=\sec x\tan x):

f'(x)=\dfrac{\sec x\tan x(\sec x+1)-(\sec x-1)\sec x\tan x}{(\sec x+1)^2}

Taking out the common factor \sec x\tan x:

=\dfrac{\sec x\tan x\left[(\sec x+1)-(\sec x-1)\right]}{(\sec x+1)^2}=\dfrac{2\sec x\tan x}{(\sec x+1)^2}

Using the power rule together with the chain rule (the inner function is \sin x, with derivative \cos x):

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

Using the quotient rule with u=a+b\sin x (u'=b\cos x) and v=c+d\cos x (v'=-d\sin x):

f'(x)=\dfrac{b\cos x(c+d\cos x)-(a+b\sin x)(-d\sin x)}{(c+d\cos x)^2}

Expanding the numerator:

bc\cos x+bd\cos^2x+ad\sin x+bd\sin^2x=bc\cos x+bd\left(\cos^2x+\sin^2x\right)+ad\sin x

=ad\sin x+bc\cos x+bd

Using the quotient rule with u=\sin(x+a) (u'=\cos(x+a)) and v=\cos x (v'=-\sin x):

f'(x)=\dfrac{\cos(x+a)\cos x-\sin(x+a)(-\sin x)}{\cos^2x}=\dfrac{\cos(x+a)\cos x+\sin(x+a)\sin x}{\cos^2x}

Using \cos A\cos B+\sin A\sin B=\cos(A-B) with A=x+a, B=x:

=\dfrac{\cos\left[(x+a)-x\right]}{\cos^2x}=\dfrac{\cos a}{\cos^2x}

Using the product rule with u=x^4 (u'=4x^3) and v=5\sin x-3\cos x (v'=5\cos x+3\sin x):

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

=20x^3\sin x-12x^3\cos x+5x^4\cos x+3x^4\sin x

Using the product rule with u=x^2+1 (u'=2x) and v=\cos x (v'=-\sin x):

f'(x)=2x\cos x+(x^2+1)(-\sin x)=2x\cos x-(x^2+1)\sin x

Using the product rule with u=ax^2+\sin x (u'=2ax+\cos x) and v=p+q\cos x (v'=-q\sin x):

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

Expanding:

=2apx+2aqx\cos x+p\cos x+q\cos^2x-aqx^2\sin x-q\sin^2x

Using \cos^2x-\sin^2x=\cos2x:

=2apx+2aqx\cos x-aqx^2\sin x+p\cos x+q\cos2x

Using the product rule with u=x+\cos x (u'=1-\sin x) and v=x-\tan x (v'=1-\sec^2x=-\tan^2x):

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

Expanding the first product:

(1-\sin x)(x-\tan x)=x-\tan x-x\sin x+\sin x\tan x

Note that \sin x\tan x=\dfrac{\sin^2x}{\cos x} and, from the second product, \cos x\tan^2x=\dfrac{\sin^2x}{\cos x} — these two terms are equal and cancel when combined:

f'(x)=x-\tan x-x\sin x+\dfrac{\sin^2x}{\cos x}-x\tan^2x-\dfrac{\sin^2x}{\cos x}=x-x\sin x-x\tan^2x-\tan x

Using the quotient rule with u=4x+5\sin x (u'=4+5\cos x) and v=3x+7\cos x (v'=3-7\sin x):

f'(x)=\dfrac{(4+5\cos x)(3x+7\cos x)-(4x+5\sin x)(3-7\sin x)}{(3x+7\cos x)^2}

Expanding both products in the numerator:

(4+5\cos x)(3x+7\cos x)=12x+28\cos x+15x\cos x+35\cos^2x

(4x+5\sin x)(3-7\sin x)=12x-28x\sin x+15\sin x-35\sin^2x

Subtracting, the 12x terms cancel and 35\cos^2x+35\sin^2x=35:

=28\cos x+15x\cos x+35+28x\sin x-15\sin x

Since \cos\dfrac{\pi}{4}=\dfrac{1}{\sqrt2} is a constant, write f(x)=\dfrac{1}{\sqrt2}\cdot\dfrac{x^2}{\sin x}.

Using the quotient rule on \dfrac{x^2}{\sin x} with u=x^2 (u'=2x) and v=\sin x (v'=\cos x):

\dfrac{d}{dx}\left(\dfrac{x^2}{\sin x}\right)=\dfrac{2x\sin x-x^2\cos x}{\sin^2x}

So:

f'(x)=\dfrac{1}{\sqrt2}\times\dfrac{2x\sin x-x^2\cos x}{\sin^2x}=\dfrac{x(2\sin x-x\cos x)}{\sqrt2\sin^2x}

Using the quotient rule with u=x (u'=1) and v=1+\tan x (v'=\sec^2x):

f'(x)=\dfrac{1(1+\tan x)-x\sec^2x}{(1+\tan x)^2}

Using the product rule with u=x+\sec x (u'=1+\sec x\tan x) and v=x-\tan x (v'=1-\sec^2x=-\tan^2x):

f'(x)=(1+\sec x\tan x)(x-\tan x)+(x+\sec x)(-\tan^2x)

Expanding the first product:

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

and the second product:

(x+\sec x)(-\tan^2x)=-x\tan^2x-\sec x\tan^2x

Adding these together:

f'(x)=x-\tan x+x\sec x\tan x-\sec x\tan^2x-x\tan^2x-\sec x\tan^2x

=x\left(1+\sec x\tan x-\tan^2x\right)-\tan x-2\sec x\tan^2x

Write f(x)=x(\sin x)^{-n} and use the product rule with u=x (u'=1) and v=(\sin x)^{-n} (using the chain rule, v'=-n(\sin x)^{-n-1}\cos x):

f'(x)=(\sin x)^{-n}+x\times\left[-n(\sin x)^{-n-1}\cos x\right]=(\sin x)^{-n}-nx\cos x(\sin x)^{-n-1}

Taking out the common factor (\sin x)^{-n-1} (using (\sin x)^{-n}=\sin x\cdot(\sin x)^{-n-1}):

f'(x)=(\sin x)^{-n-1}\left[\sin x-nx\cos x\right]=\dfrac{\sin x-nx\cos x}{\sin^{n+1}x}