Differentiation Class 12 Maths 2 Exercise 1.2 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Ex 1.2 Questions and Answers.

12th Maths Part 2 Differentiation Exercise 1.2 Questions And Answers Maharashtra Board

Question 1.
Find the derivative of the function y = f(x) using the derivative of the inverse function x = f -1 ( y) in the following
(i) y = \(\sqrt {x}\)
Solution:
y = \(\sqrt {x}\) … (1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),
y 2 = x ∴ x = y 2
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-1

(ii) y = \(\sqrt{2-\sqrt{x}}\)
Solution:
y = \(\sqrt{2-\sqrt{x}}\) …(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-2

(iii) y = \(\sqrt[3]{x-2}\)
Solution:
y = \(\sqrt[3]{x-2}\) ….(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-3
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-4

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(iv) y = log (2x – 1)
Solution:
y = log (2x – 1) …(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-5

(v) y = 2x + 3
Solution:
y = 2x + 3 ….(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-6

(vi) y = e x – 3
Solution:
y = e x – 3 ….(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),
e x = y + 3
∴ x = log(y + 3)
∴ x = f -1 (y) = log(y + 3)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-7

Maharashtra-Board-Solutions

(vii) y = e 2x – 3
Solution:
y = e 2x – 3 ….(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),
2x – 3 = log y ∴ 2x = log y + 3
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-8

(viii) y = log 2 \(\left(\frac{x}{2}\right)\)
Solution:
y = log 2 \(\left(\frac{x}{2}\right)\) …(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),
\(\frac{x}{2}\) = 2 y ∴ x = 2βˆ™2 y = 2 y+1
∴ x = f -1 (y) = 2 y+1
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-9

Question 2.
Find the derivative of the inverse function of
the following
(i) y = x 2 Β·e x
Solution:
y = x 2 Β·e x
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-10

Maharashtra-Board-Solutions

(ii) y = x cos x
Solution:
y = x cos x
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-11
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-12

(iii) y = xΒ·7 x
Solution:
y = xΒ·7 x
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-13

(iv) y = x 2 + logx
Solution:
y = x 2 + logx
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-14

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(v) y = x logx
Solution:
y = x logx
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-15
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-16

Question 3.
Find the derivative of the inverse of the following functions, and also fid their value at the points indicated against them.
(i) y = x 5 + 2x 3 + 3x, at x = 1
Solution:
y = x 5 + 2x 3 + 3x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(x 5 + 2x 3 + 3x)
= 5x 4 + 2 Γ— 3x 2 + 3 Γ— 1
= 5x 4 + 6x 2 + 3
The derivative of inverse function of y = f(x) is given by
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-17

(ii) y = e x + 3x + 2, at x = 0
Solution:
y = e x + 3x + 2
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(e x + 3x + 2)
The derivative of inverse function of y = f(x) is given by
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-18
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-19

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(iii) y = 3x 2 + 2 log x 3 , at x = 1
Solution:
y = 3x 2 + 2 log x 3
= 3x 2 + 6 log x
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-20
The derivative of inverse function of y = f(x) is given by
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-21

(iv) y = sin (x – 2) + x 2 , at x = 2
Solution:
y = sin (x – 2) + x 2
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-22
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-23

Question 4.
If f(x) = x 3 + x – 2, find (f -1 )’ (0).
Question is modified.
If f(x) = x 3 + x – 2, find (f -1 )’ (-2).
Solution:
f(x) = x 3 + x – 2 ….(1)
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-24

Maharashtra-Board-Solutions

Question 5.
Using derivative prove
(i) tan -1 x + cot -1 x = \(\frac{\pi}{2}\)
Solution:
let f(x) = tan -1 x + cot -1 x
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-25
Since, f'(x) = 0, f(x) is a constant function.
Let f(x) = k.
For any value of x, f(x) = k
Let x = 0.
Then f(0) = k ….(2)
From (1), f(0) = tan -1 (0) + cot -1 (0)
= 0 + \(\frac{\pi}{2}=\frac{\pi}{2}\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-26

(ii) sec -1 x + cosec -1 x = \(\frac{\pi}{2}\) . . . [for |x| β‰₯ 1]
Solution:
Let f(x) = sec -1 x + cosec -1 x for |x| β‰₯ 1 ….(1)
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-27
Since, f'(x) = 0, f(x) is a constant function.
Let f(x) = k.
For any value of x, f(x) = k, where |x| > 1
Let x = 2.
Then, f(2) = k ……(2)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-28

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Question 6.
Diffrentiate the following w. r. t. x.
(i) tan -1 (log x)
Solution:
Let y = tan -1 (log x)
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-29

(ii) cosec -1 (e -x )
Solution:
Let y = cosec -1 (e -x )
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-104

(iii) cot -1 (x 3 )
Solution:
Let y = cot -1 (x 3 )
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-105

Maharashtra-Board-Solutions

(iv) cot -1 (4 x
Solution:
Let y = cot -1 (4 x
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-106

(v) tan -1 (\(\sqrt {x}\))
Solution:
Let y = tan -1 (\(\sqrt {x}\))
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-107

(vi) sin -1 \(\left(\sqrt{\frac{1+x^{2}}{2}}\right)\)
Solution:
Let y = sin -1 \(\left(\sqrt{\frac{1+x^{2}}{2}}\right)\)
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-108

Maharashtra-Board-Solutions

(vii) cos -1 (1 – x 2 )
Solution:
Let y = cos -1 (1 – x 2 )
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-109
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-30

(viii) sin -1 \(\left(x^{\frac{3}{2}}\right)\)
Solution:
Let y = sin -1 \(\left(x^{\frac{3}{2}}\right)\)
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-31

Maharashtra-Board-Solutions

(ix) cos 3 [cos -1 (x 3 )]
Solution:
Let y = cos 3 [cos -1 (x 3 )]
= [cos(cos -1 x 3 )] 3
= (x 3 ) 3 = x 9
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(x 9 ) = 9x 8 .

(x) sin 4 [sin -1 (\(\sqrt {x}\))]
Solution:
Let y = sin 4 [sin -1 (\(\sqrt {x}\))]
= {sin[sin -1 (\(\sqrt {x}\))]} 8
= (\(\sqrt {x}\)) 4 = x 2
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(x 2 ) = 2x.

Question 7.
Diffrentiate the following w. r. t. x.
(i) cot -1 [cot (e x2 )]
Solution:
Let y = cot -1 [cot (e x2 )] = e x2
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-32

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(ii) cosec -1 \(\left(\frac{1}{\cos \left(5^{x}\right)}\right)\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-33

(iii) cos -1 \(\left(\sqrt{\frac{1+\cos x}{2}}\right)\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-34

(iv) cos -1 \(\left(\sqrt{\frac{1-\cos \left(x^{2}\right)}{2}}\right)\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-35
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-36

(v) tan -1 \(\left(\frac{1-\tan \left(\frac{x}{2}\right)}{1+\tan \left(\frac{x}{2}\right)}\right)\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-37
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-38

Maharashtra-Board-Solutions

(vi) cosec -1 \(\left(\frac{1}{4 \cos ^{3} 2 x-3 \cos 2 x}\right)\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-39

(vii) tan -1 \(\left(\frac{1+\cos \left(\frac{x}{3}\right)}{\sin \left(\frac{x}{3}\right)}\right)\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-40
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-41

(viii) cot -1 \(\left(\frac{\sin 3 x}{1+\cos 3 x}\right)\)
Solution:
Let y = cot -1 \(\left(\frac{\sin 3 x}{1+\cos 3 x}\right)\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-42

Maharashtra-Board-Solutions

(ix) tan -1 \(\left(\frac{\cos 7 x}{1+\sin 7 x}\right)\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-43
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-44

(x) tan -1 \(\left(\sqrt{\frac{1+\cos x}{1-\cos x}}\right)\)
Solution:
Let y = tan -1 \(\left(\sqrt{\frac{1+\cos x}{1-\cos x}}\right)\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-45

(xi) tan -1 (cosec x + cot x)
Solution:
Let y = tan -1 (cosec x + cot x)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-46
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-47

Maharashtra-Board-Solutions

(xii) cot -1 \(\left(\frac{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}+\sqrt{1-\sin \left(\frac{4 x}{3}\right)}}{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}-\sqrt{1-\sin \left(\frac{4 x}{3}\right)}}\right)\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-48
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-49
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-50
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-51

Question 8.
(i) Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-60
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-52
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-53

(ii) Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-61
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-54
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-55

Maharashtra-Board-Solutions

(iii) Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-62
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-56
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-57

(iv) Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-63
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-58
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-59

(v) Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-64
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-65
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-66
= e x .

Maharashtra-Board-Solutions

(vi) Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-67
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-68
y = sin -1 [sin(2 x )βˆ™cosΞ± – cos(2 x )βˆ™sinΞ±]
= sin – [sin(2 x – Ξ±)]
= 2 x – Ξ±, where Ξ± is a constant
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(2 x – Ξ±)
= \(\frac{d}{d x}\)(2 x ) – \(\frac{d}{d x}\)(Ξ±)
= 2 x βˆ™log2 – 0
= 2 x βˆ™log2

Question 9.
Diffrentiate the following w. r. t. x.
(i) cos -1 \(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-69

(ii) tan -1 \(\left(\frac{2 x}{1-x^{2}}\right)\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-70

Maharashtra-Board-Solutions

(iii) sin -1 \(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-71
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-72

(iv) sin -1 (2x\(\sqrt{1-x^{2}}\))
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-73
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-74

(v) cos -1 (3x – 4x 3 )
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-75
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-76

(vi) cos -1 \(\left(\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right)\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-77
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-78

Maharashtra-Board-Solutions

(vii) cos -1 \(\left(\frac{1-9^{x}}{1+9^{x}}\right)\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-79
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-80

(viii) sin -1 \(\left(\frac{4^{x+\frac{1}{2}}}{1+2^{4 x}}\right)\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-81
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-82

(ix) sin -1 \(\left(\frac{1-25 x^{2}}{1+25 x^{2}}\right)\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-83
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-84

(x) sin -1 \(\left(\frac{1-x^{3}}{1+x^{3}}\right)\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-85
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-86

Maharashtra-Board-Solutions

(xi) tan -1 \(\left(\frac{2 x^{\frac{5}{2}}}{1-x^{5}}\right)\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-87
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-88

(xii) cot -1 \(\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)\)
Solution:
Let y = cot -1 \(\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-89
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-90

Question 10.
Diffrentiate the following w. r. t. x.
(i) tan -1 \(\left(\frac{8 x}{1-15 x^{2}}\right)\)
Solution:
Let y = tan -1 \(\left(\frac{8 x}{1-15 x^{2}}\right)\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-91

Maharashtra-Board-Solutions

(ii) cot -1 \(\left(\frac{1+35 x^{2}}{2 x}\right)\)
Solution:
Let y = cot -1 \(\left(\frac{1+35 x^{2}}{2 x}\right)\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-92
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-93

(iii) tan -1 \(\left(\frac{2 \sqrt{x}}{1+3 x}\right)\)
Solution:
Let y = tan -1 \(\left(\frac{2 \sqrt{x}}{1+3 x}\right)\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-94

(iv) tan -1 \(\left(\frac{2^{x+2}}{1-3\left(4^{x}\right)}\right)\)
Solution:
Let y = tan -1 \(\left(\frac{2^{x+2}}{1-3\left(4^{x}\right)}\right)\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-95
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-96

Maharashtra-Board-Solutions

(v) tan -1 \(\left(\frac{2^{x}}{1+2^{2 x+1}}\right)\)
Solution:
Let y = tan -1 \(\left(\frac{2^{x}}{1+2^{2 x+1}}\right)\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-97

(vi) cot -1 \(\left(\frac{a^{2}-6 x^{2}}{5 a x}\right)\)
Solution:
Let y = cot -1 \(\left(\frac{a^{2}-6 x^{2}}{5 a x}\right)\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-98
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-99

(vii) tan -1 \(\left(\frac{a+b \tan x}{b-a \tan x}\right)\)
Solution:
Let y = tan -1 \(\left(\frac{a+b \tan x}{b-a \tan x}\right)\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-100

Maharashtra-Board-Solutions

(viii) tan -1 \(\left(\frac{5-x}{6 x^{2}-5 x-3}\right)\)
Solution:
Let y = tan -1 \(\left(\frac{5-x}{6 x^{2}-5 x-3}\right)\)
= tan -1 \(\left[\frac{5-x}{1+\left(6 x^{2}-5 x-4\right)}\right]\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-101

(ix) cot -1 \(\left(\frac{4-x-2 x^{2}}{3 x+2}\right)\)
Solution:
Let y = cot -1 \(\left(\frac{4-x-2 x^{2}}{3 x+2}\right)\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-102
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.2-103