NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2 - Limits and Derivatives

Question1: Find the derivative of $x^2-2atx=10$

Answer:

F(x)= $x^2-2$

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

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$

The derivative of f(x) at x = 10:

$f^{\prime }(10)=\underset{h\to 0}{lim}\frac{f(10+h)-f(10)}{h}$

$f^{\prime }(10)=\underset{h\to 0}{lim}\frac{(10+h{)}^2-2-((10{)}^2-2)}{h}$

$f^{\prime }(10)=\underset{h\to 0}{lim}\frac{100+20h+h^2-2-100+2}{h}$

$f^{\prime }(10)=\underset{h\to 0}{lim}\frac{20h+h^2}{h}$

$f^{\prime }(10)=\underset{h\to 0}{lim}20+h$

$f^{\prime }(10)=20+0$

$f^{\prime }(10)=20$

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

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$

The derivative of f(x) at x = 1:

$f^{\prime }(1)=\underset{h\to 0}{lim}\frac{f(1+h)-f(1)}{h}$

$f^{\prime }(1)=\underset{h\to 0}{lim}\frac{(1+h)-(1)}{h}$

$f^{\prime }(1)=\underset{h\to 0}{lim}\frac{(h)}{h}$

$f^{\prime }(1)=1$ (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

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$

The derivative of f(x) at x = 100:

$f^{\prime }(100)=\underset{h\to 0}{lim}\frac{f(100+h)-f(100)}{h}$

$f^{\prime }(100)=\underset{h\to 0}{lim}\frac{99(100+h)-99(100)}{h}$

$f^{\prime }(100)=\underset{h\to 0}{lim}\frac{99h}{h}$

$f^{\prime }(100)=99$

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

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{(x+h{)}^3-27-((x{)}^3-27)}{h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{x^3+h^3+3x^{2}h+3h^{2}x-27+x^3+27}{h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{h^3+3x^{2}h+3h^{2}x}{h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}{h}^2+3x^2+3hx$

$f^{\prime }(x)=3x^2$

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

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{(x+h-1)(x+h-2)-(x-1)(x-2)}{h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{x^2+xh-2x+hx+h^2-2h-x-h+2-x^2+2x+x-2}{h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{2hx+h^2-3h}{h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}2x+h-3$

$f^{\prime }(x)=2x-3$ (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

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{1/(x+h{)}^2-1/(x^2)}{h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{\frac{x^2-(x+h{)}^2}{(x+h{)}^2{x}^2}}{h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{x^2-x^2-2xh-h^2}{h(x+h{)}^2{x}^2}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{-2xh-h^2}{h(x+h{)}^2{x}^2}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{-2x-h}{(x+h{)}^2{x}^2}$

$f^{\prime }(x)=\frac{-2x-0}{(x+0{)}^2{x}^2}$

$f^{\prime }(x)=\frac{-2}{x^3}$ (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

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{\frac{x+h+1}{x+h-1}-\frac{x+1}{x-1}}{h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{\frac{(x+h+1)(x-1)-(x+1)(x+h-1)}{(x-1)(x+h-1)}}{h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{x^2-x+hx-h+x-1-x^2-xh+x-x-h+1}{(x-1)(x+h-1)h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{-2h}{(x-1)(x+h-1)h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{-2}{(x-1)(x+h-1)}$

$f^{\prime }(x)=\frac{-2}{(x-1)(x+0-1)}$

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

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,

$f^{\prime }(x^n)=n{x}^{n-1}$

applying that property we get

$f^{\prime }(x)=100\frac{x^{99}}{100}+99\frac{x^{98}}{99}+....+2\frac{x}{2}+1+0$

$f^{\prime }(x)=x^{99}+x^{98}+......x+1$

Now.

$f^{\prime }(0)=0^{99}+0^{98}+......0+1=1$

$f^{\prime }(1)=1^{99}+1^{98}+......1+1=100$

So,

Here

$1\times 100=100$

$f^{\prime }(0)\times 100=f^{\prime }(1)$

Hence Proved.

Question 6: Find the derivative of $x^n+a{x}^{n-1}+a^2{x}^{n-2}+....+a^{n-1}x+a^n$ for some fixed real number a.

Answer:

Given

$f(x)=x^n+a{x}^{n-1}+a^2{x}^{n-2}+....+a^{n-1}x+a^n$

As we know, the property,

$f^{\prime }(x^n)=n{x}^{n-1}$

applying that property we get

$f^{\prime }(x)=n{x}^{n-1}+a(n-1)x^{n-2}+a^2(n-2)x^{n-3}+....+a^{n-1}1+0$

$f^{\prime }(x)=n{x}^{n-1}+a(n-1)x^{n-2}+a^2(n-2)x^{n-3}+....+a^{n-1}$

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,

$f^{\prime }(x^n)=n{x}^{n-1}$

and the property

$\frac{d(y_1+y_2)}{dx}=\frac{d{y}_1}{dx}+\frac{d{y}_2}{dx}$

applying that property we get

$f^{\prime }(x)=2x-a-b$

Question 7:(ii) For some constants a and b, find the derivative of $(a{x}^2+b{)}^2$

Answer:

Given

$f(x)=(a{x}^2+b{)}^2=a^2{x}^4+2ab{x}^2+b^2$

As we know, the property,

$f^{\prime }(x^n)=n{x}^{n-1}$

and the property

$\frac{d(y_1+y_2)}{dx}=\frac{d{y}_1}{dx}+\frac{d{y}_2}{dx}$

applying those properties we get

$f^{\prime }(x)=4a^2{x}^3+2(2)abx+0$

$f^{\prime }(x)=4a^2{x}^3+4abx$

$f^{\prime }(x)=4ax(a{x}^2+b)$

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,

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{d{y}_1}{dx}-y_1\frac{d{y}_2}{dx}}{y_2^2}$

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

$f^{\prime }(x)=\frac{a-b}{(x-b{)}^2}$

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,

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{d{y}_1}{dx}-y_1\frac{d{y}_2}{dx}}{y_2^2}$

So applying this rule, we get

$\frac{d(\frac{x^n-a^n}{x-a})}{dx}=\frac{(x-a)\frac{d(x^n-a^n)}{dx}-(x^n-a^n)\frac{d(x-a)}{dx}}{(x-a{)}^2}$

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

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

Hence

$f^{\prime }(x)=\frac{n{x}^n-an{x}^{n-1}-x^n+a^n}{(x-a{)}^2}$

Question 9:(i) Find the derivative of $2x-3/4$

Answer:

Given:

$f(x)=2x-3/4$

As we know, the property,

$f^{\prime }(x^n)=n{x}^{n-1}$

and the property

$\frac{d(y_1+y_2)}{dx}=\frac{d{y}_1}{dx}+\frac{d{y}_2}{dx}$

applying that property we get

$f^{\prime }(x)=2-0$

$f^{\prime }(x)=2$

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,

$f^{\prime }(x^n)=n{x}^{n-1}$

and the property

$\frac{d(y_1+y_2)}{dx}=\frac{d{y}_1}{dx}+\frac{d{y}_2}{dx}$

applying that property we get

$f^{\prime }(x)=5(4)x^3-5(3)x^2+3(2)x-4+0$

$f^{\prime }(x)=20x^3-15x^2+6x-4$

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,

$f^{\prime }(x^n)=n{x}^{n-1}$

and the property

$\frac{d(y_1+y_2)}{dx}=\frac{d{y}_1}{dx}+\frac{d{y}_2}{dx}$

applying that property we get

$f^{\prime }(x)=(-3)5x^{-4}+3(-2)x^{-3}$

$f^{\prime }(x)=-15x^{-4}-6x^{-3}$

Question 9:(iv) Find the derivative of $x^5(3-6x^{-9})$

Answer:

Given

$f(x)=x^5(3-6x^{-9})=3x^5-6x^{-4}$

As we know, the property,

$f^{\prime }(x^n)=n{x}^{n-1}$

and the property

$\frac{d(y_1+y_2)}{dx}=\frac{d{y}_1}{dx}+\frac{d{y}_2}{dx}$

applying that property we get

$f^{\prime }(x)=(5)3x^4-6(-4)x^{-5}$

$f^{\prime }(x)=15x^4+24x^{-5}$

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,

$f^{\prime }(x^n)=n{x}^{n-1}$

and the property

$\frac{d(y_1+y_2)}{dx}=\frac{d{y}_1}{dx}+\frac{d{y}_2}{dx}$

applying that property we get

$f^{\prime }(x)=(-4)3x^{-5}-(-9)4x^{-10}$

$f^{\prime }(x)=-12x^{-5}+36x^{-10}$

Question 9:(vi) Find the derivative of $\frac{2}{x+1}-\frac{x^2}{3x-1}$

Answer:

Given

$f(x)=\frac{2}{x+1}-\frac{x^2}{3x-1}$

As we know the quotient rule of derivative:

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{d{y}_1}{dx}-y_1\frac{d{y}_2}{dx}}{y_2^2}$

and the property

$\frac{d(y_1+y_2)}{dx}=\frac{d{y}_1}{dx}+\frac{d{y}_2}{dx}$

So applying this rule, we get

$\frac{d(\frac{2}{x+1}-\frac{x^2}{3x-1})}{dx}=\frac{(x+1)\frac{d(2)}{dx}-2\frac{d(x+1)}{dx}}{(x+1{)}^2}-\frac{(3x-1)\frac{d(x^2)}{dx}-x^2\frac{d(3x-1)}{dx}}{(3x-1{)}^2}$

$\frac{d(\frac{2}{x+1}-\frac{x^2}{3x-1})}{dx}=\frac{-2}{(x+1{)}^2}-\frac{(3x-1)2x-x^{2}3}{(3x-1{)}^2}$

$\frac{d(\frac{2}{x+1}-\frac{x^2}{3x-1})}{dx}=\frac{-2}{(x+1{)}^2}-\frac{6x^2-2x-3x^2}{(3x-1{)}^2}$

$\frac{d(\frac{2}{x+1}-\frac{x^2}{3x-1})}{dx}=\frac{-2}{(x+1{)}^2}-\frac{3x^2-2x}{(3x-1{)}^2}$

Hence

$f^{\prime }(x)=\frac{-2}{(x+1{)}^2}-\frac{3x^2-2x}{(3x-1{)}^2}$

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

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{\cos (x+h)-\cos (x)}{h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{\cos (x)\cos (h)-\sin (x)\sin (h)-\cos (x)}{h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{\cos (x)\cos (h)-\cos (x)}{h}-\frac{\sin (x)\sin (h)}{h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{\cos (x)(\cos (h)-1)}{h}-\frac{\sin (x)\sin (h)}{h}$

$f^{\prime }(x)=\underset{h\to 0}{lim}\cos (x)\frac{-2si{n}^2(h/2)}{h}\cdot -\sin (x)\frac{sinh}{h}$

$f^{\prime }(x)=\cos (x)(0)-sinx(1)$

$f^{\prime }(x)=-\sin (x)$

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,

$\frac{d(y_1{y}_2)}{dx}=y_1\frac{d{y}_2}{dx}+y_2\frac{d{y}_1}{dx}$

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,

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{d{y}_1}{dx}-y_1\frac{d{y}_2}{dx}}{y_2^2}$

So applying this rule, we get

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

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

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

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

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

$\frac{d(y_1+y_2)}{dx}=\frac{d{y}_1}{dx}+\frac{d{y}_2}{dx}$

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,

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{d{y}_1}{dx}-y_1\frac{d{y}_2}{dx}}{y_2^2}$

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

$\frac{d(y_1+y_2)}{dx}=\frac{d{y}_1}{dx}+\frac{d{y}_2}{dx}$

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,

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{d{y}_1}{dx}-y_1\frac{d{y}_2}{dx}}{y_2^2}$

So applying this rule, we get

$\frac{d(3\cot x+5\csc x)}{dx}=-5\csc x\cot x+3\left[\frac{\sin x\frac{d(\cos x)}{dx}-\cos x(\frac{d(\sin x)}{dx})}{{\sin }^{2}x}\right]$

$\frac{d(3\cot x+5\csc x)}{dx}=-5\csc x\cot x+3\left[\frac{\sin x(-\sin x)-\cos x(\cos x)}{{\sin }^{2}x}\right]$

$\frac{d(3\cot x+5\csc x)}{dx}=-5\csc x\cot x+3\left[\frac{-{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{2}x}\right]$

$\frac{d(3\cot x+5\csc x)}{dx}=-5\csc x\cot x-3\left[\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x}\right]$

$\frac{d(3\cot x+5\csc x)}{dx}=-5\csc x\cot x-3\left[\frac{1}{{\sin }^{2}x}\right]$

$\frac{d(3\cot x+5\csc x)}{dx}=-5\csc x\cot x-3{\csc }^{2}x$