Class 11 Maths NCERT Solutions Chapter 12 Ex 12.2 – Limits and Derivatives

Using \dfrac{d}{dx}(x^n)=nx^{n-1} along with the sum rule:

\dfrac{d}{dx}(x^2-2)=2x-0=2x

At x=10:

f'(10)=2(10)=20

The derivative of f(x)=x is the constant function 1, since \dfrac{d}{dx}(x)=1\cdot x^{1-1}=1.

This value is 1 regardless of the point, so:

f'(1)=1

Writing f(x)=99x as the product of the constant 99 and x:

\dfrac{d}{dx}(99x)=99\times\dfrac{d}{dx}(x)=99\times1=99

This value does not depend on x, so it is the same at x=100:

f'(100)=99

By definition, the derivative from first principle is f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}.

f(x+h)-f(x)=\left[(x+h)^3-27\right]-\left[x^3-27\right]=(x+h)^3-x^3

=x^3+3x^2h+3xh^2+h^3-x^3=h\left(3x^2+3xh+h^2\right)

f'(x)=\lim_{h\to0}\left(3x^2+3xh+h^2\right)=3x^2

Expanding, f(x)=x^2-3x+2, so:

f(x+h)-f(x)=\left[(x+h)^2-3(x+h)+2\right]-\left[x^2-3x+2\right]=2xh+h^2-3h=h(2x+h-3)

f'(x)=\lim_{h\to0}(2x+h-3)=2x-3

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

f'(x)=\lim_{h\to0}\dfrac{-(2x+h)}{x^2(x+h)^2}=\dfrac{-2x}{x^2\cdot x^2}=-\dfrac{2}{x^3}

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

Expanding the numerator: (x+h+1)(x-1)=x^2+hx-h-1 and (x+1)(x+h-1)=x^2+hx+h-1, so the numerator simplifies to -2h.

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

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

Differentiating term by term using \dfrac{d}{dx}\left(\dfrac{x^k}{k}\right)=x^{k-1} for each of the 100 terms (with the constant 1 contributing 0):

f'(x)=x^{99}+x^{98}+\ldots+x^2+x+1

This is a sum of 100 terms, x^{99},x^{98},\ldots,x,1.

At x=1, every term equals 1, so the sum of 100 terms is:

f'(1)=1+1+\ldots+1\ (100\text{ terms})=100

At x=0, every term with a positive power of x vanishes, leaving only the constant term:

f'(0)=0+0+\ldots+0+1=1

Therefore 100f'(0)=100(1)=100=f'(1).

Here, a is a fixed constant and x is the variable. The general term is a^kx^{n-k} for k=0,1,\ldots,n, so each term is differentiated using \dfrac{d}{dx}\left(a^kx^{n-k}\right)=a^k(n-k)x^{n-k-1} (the last term, a^n, is constant and its derivative is 0).

Differentiating term by term:

\dfrac{d}{dx}\left(x^n+ax^{n-1}+a^2x^{n-2}+\ldots+a^{n-1}x+a^n\right)

=nx^{n-1}+(n-1)ax^{n-2}+(n-2)a^2x^{n-3}+\ldots+2a^{n-2}x+a^{n-1}

Expanding, f(x)=x^2-(a+b)x+ab, so:

f'(x)=2x-(a+b)

Using the product rule with u=v=ax^2+b (equivalently, the chain rule with the outer power 2):

f'(x)=2(ax^2+b)\cdot\dfrac{d}{dx}(ax^2+b)=2(ax^2+b)(2ax)

=4a^2x^3+4abx

Using the quotient rule with u=x-a, v=x-b, so u'=1 and v'=1:

f'(x)=\dfrac{u'v-uv'}{v^2}=\dfrac{1(x-b)-(x-a)(1)}{(x-b)^2}=\dfrac{x-b-x+a}{(x-b)^2}

=\dfrac{a-b}{(x-b)^2}

Using the quotient rule with u=x^n-a^n, v=x-a, so u'=nx^{n-1} and v'=1:

f'(x)=\dfrac{u'v-uv'}{v^2}=\dfrac{nx^{n-1}(x-a)-(x^n-a^n)(1)}{(x-a)^2}

Expanding the numerator:

nx^{n-1}(x-a)-(x^n-a^n)=nx^n-nax^{n-1}-x^n+a^n=(n-1)x^n-nax^{n-1}+a^n

The term -\dfrac{3}{4} is constant, so its derivative is 0:

f'(x)=2

Using the product rule with u=5x^3+3x-1 (so u'=15x^2+3) and v=x-1 (so v'=1):

f'(x)=u'v+uv'=(15x^2+3)(x-1)+(5x^3+3x-1)(1)

=15x^3-15x^2+3x-3+5x^3+3x-1

=20x^3-15x^2+6x-4

Expanding first: f(x)=5x^{-3}+3x^{-2}.

f'(x)=-15x^{-4}-6x^{-3}=-\dfrac{15}{x^4}-\dfrac{6}{x^3}

Expanding first: f(x)=3x^5-6x^{-4}.

f'(x)=15x^4+24x^{-5}=15x^4+\dfrac{24}{x^5}

Expanding first: f(x)=3x^{-4}-4x^{-9}.

f'(x)=-12x^{-5}+36x^{-10}=-\dfrac{12}{x^5}+\dfrac{36}{x^{10}}

Differentiate each term separately using the quotient rule.

For \dfrac{2}{x+1}: with u=2 (constant, u'=0) and v=x+1 (v'=1):

\dfrac{d}{dx}\left(\dfrac{2}{x+1}\right)=\dfrac{0(x+1)-2(1)}{(x+1)^2}=-\dfrac{2}{(x+1)^2}

For \dfrac{x^2}{3x-1}: with u=x^2 (u'=2x) and v=3x-1 (v'=3):

\dfrac{d}{dx}\left(\dfrac{x^2}{3x-1}\right)=\dfrac{2x(3x-1)-x^2(3)}{(3x-1)^2}=\dfrac{6x^2-2x-3x^2}{(3x-1)^2}=\dfrac{3x^2-2x}{(3x-1)^2}

Subtracting:

f'(x)=-\dfrac{2}{(x+1)^2}-\dfrac{3x^2-2x}{(3x-1)^2}

Let f(x)=\cos x. By definition:

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

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

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

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

=-\lim_{h\to0}\sin\left(x+\dfrac{h}{2}\right)\times\lim_{h\to0}\dfrac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}=-\sin x\times1=-\sin x

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

f'(x)=\cos x\cos x+\sin x(-\sin x)=\cos^2x-\sin^2x=\cos2x

Write \sec x=\dfrac{1}{\cos x}. Using the quotient rule with u=1 (u'=0) and v=\cos x (v'=-\sin x):

f'(x)=\dfrac{0(\cos x)-1(-\sin x)}{\cos^2x}=\dfrac{\sin x}{\cos^2x}=\dfrac{1}{\cos x}\cdot\dfrac{\sin x}{\cos x}=\sec x\tan x

Using the sum rule with the results \dfrac{d}{dx}(\sec x)=\sec x\tan x and \dfrac{d}{dx}(\cos x)=-\sin x:

f'(x)=5\sec x\tan x+4(-\sin x)=5\sec x\tan x-4\sin x

Write \operatorname{cosec}x=\dfrac{1}{\sin x}. Using the quotient rule with u=1 (u'=0) and v=\sin x (v'=\cos x):

f'(x)=\dfrac{0(\sin x)-1(\cos x)}{\sin^2x}=\dfrac{-\cos x}{\sin^2x}=-\dfrac{1}{\sin x}\cdot\dfrac{\cos x}{\sin x}=-\operatorname{cosec}x\cot x

Using the sum rule with the standard results \dfrac{d}{dx}(\cot x)=-\operatorname{cosec}^2x and \dfrac{d}{dx}(\operatorname{cosec}x)=-\operatorname{cosec}x\cot x (found in part (iv)):

f'(x)=3(-\operatorname{cosec}^2x)+5(-\operatorname{cosec}x\cot x)=-3\operatorname{cosec}^2x-5\operatorname{cosec}x\cot x

Using the sum/difference rule, and noting the constant 7 has derivative 0:

f'(x)=5\cos x-6(-\sin x)+0=5\cos x+6\sin x

Using the standard results \dfrac{d}{dx}(\tan x)=\sec^2x and \dfrac{d}{dx}(\sec x)=\sec x\tan x (found in part (ii)):

f'(x)=2\sec^2x-7\sec x\tan x