Differentiation Class 12 Maths 2 Exercise 1.1 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Ex 1.1 Questions and Answers.

12th Maths Part 2 Differentiation Exercise 1.1 Questions And Answers Maharashtra Board

Question 1.
Differentiate the following w.r.t. x :
(i) (x 3 – 2x – 1) 5
Solution:
Method 1:
Let y = (x 3 – 2x – 1) 5
Put u = x 3 – 2x – 1. Then y = u 5
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-1
Method 2:
Let y = (x 3 – 2x – 1) 5
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-2

(ii) \(\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{5}{2}}\)
Solution:
Let y = \(\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{5}{2}}\)
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-3

(iii) \(\sqrt{x^{2}+4 x-7}\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-4

Maharashtra-Board-Solutions

(iv) \(\sqrt{x^{2}+\sqrt{x^{2}+1}}\)
Solution:
Let y = \(\sqrt{x^{2}+\sqrt{x^{2}+1}}\)
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-5
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-6

(v) \(\frac{3}{5 \sqrt[3]{\left(2 x^{2}-7 x-5\right)^{5}}}\)
Solution:
Let y = \(\frac{3}{5 \sqrt[3]{\left(2 x^{2}-7 x-5\right)^{5}}}\)
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-7

(vi) \(\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^{5}\)
Solution:
Let y = \(\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^{5}\)
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-8
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-9

Maharashtra-Board-Solutions

Question 2.
Diffrentiate the following w.r.t. x
(i) cos(x 2 + a 2 )
Solution:
Let y = cos(x 2 + a 2 )
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)[cos(x 2 + a 2 )]
= -sin(x 2 + a 2 )βˆ™\(\frac{d}{d x}\)x 2 + a 2 )
= -sin(x 2 + a 2 )βˆ™(2x + 0)
= -2xsin(x 2 + a 2 )

(ii) \(\sqrt{e^{(3 x+2)}+5}\)
Solution:
Let y = \(\sqrt{e^{(3 x+2)}+5}\)
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-10

(iii) log[tan(\(\frac{x}{2}\))]
Solution:
Let y = log[tan(\(\frac{x}{2}\))]
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-11

Maharashtra-Board-Solutions

(iv) \(\sqrt{\tan \sqrt{x}}\)
Solution:
Let y = \(\sqrt{\tan \sqrt{x}}\)
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-12

(v) cot 3 [log (x 3 )]
Solution:
Let y = cot 3 [log (x 3 )]
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-13

(vi) 5 sin3x+ 3
Solution:
Let y = 5 sin3x+ 3
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-14

Maharashtra-Board-Solutions

(vii) cosec (\(\sqrt{\cos X}\))
Solution:
Let y = cosec (\(\sqrt{\cos X}\))
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-15

(viii) log[cos (x 3 – 5)]
Solution:
Let y = log[cos (x 3 – 5)]
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-16

(ix) e 3 sin2x – 2 cos2x
Solution:
Let y = e 3 sin2x – 2 cos2x
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-17

Maharashtra-Board-Solutions

(x) cos 2 [log (x 2 + 7)]
Solution:
Let y = cos 2 [log (x 2 + 7)]
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-18
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-19

(xi) tan[cos (sinx)]
Solution:
Let y = tan[cos (sinx)]
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-20

(xii) sec[tan (x 4 + 4)]
Solution:
Let y = sec[tan (x 4 + 4)]
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-21
= sec[tan(x 4 + 4)]βˆ™tan[tan(x 4 + 4)]βˆ™sec 2 (x 4 + 4)(4x 3 + 0)
= 4x 3 sec 2 (x 4 + 4)βˆ™sec[tan(x 4 + 4)]βˆ™tan[tan(x 4 + 4)].

(xiii) e log[(logx)2 – logx2]
Solution:
Let y = e log[(logx)2 – logx2]
= (log x) 2 – log x 2 …[∡ e log x = x]
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-22

Maharashtra-Board-Solutions

(xiv) sin\(\sqrt{\sin \sqrt{x}}\)
Solution:
Let y = sin\(\sqrt{\sin \sqrt{x}}\)
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-23

(xv) log[sec(e x2 )]
Solution:
Let y = log[sec(e x2 )]
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-24

(xvi) log e 2 (logx)
Solution:
Let y = log e 2 (logx) = \(\frac{\log (\log x)}{\log e^{2}}\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-25
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-26

(xvii) [log{log(logx)}] 2
Solution:
let y = [log{log(logx)}] 2
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-27

Maharashtra-Board-Solutions

(xviii) sin 2 x 2 – cos 2 x 2
Solution:
Let y = sin 2 x 2 – cos 2 x 2
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-28
= 2sinx 2 βˆ™cosx 2 Γ— 2x + 2sinx 2 βˆ™cosx 2 Γ— 2x
= 4x(2sinx 2 βˆ™cosx 2 )
= 4xsin(2x 2 ).

Question 3.
Diffrentiate the following w.r.t. x
(i) (x 2 + 4x + 1) 3 + (x 3 – 5x – 2) 4
Solution:
Let y = (x 2 + 4x + 1) 3 + (x 3 – 5x – 2) 4
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)[(x 2 + 4x + 1) 3 + (x 3 – 5x – 2) 4 ]
= \(\frac{d}{d x}\) = (x 2 + 4x + 1) 3 + \(\frac{d}{d x}\)(x 3 – 5x – 2) 4
= 3(x 2 + 4x + 1) 2 βˆ™\(\frac{d}{d x}\)(x 2 + 4x + 1) + 4(x 3 – 5x – 2) 4 βˆ™\(\frac{d}{d x}\)(x 3 – 5x – 2)
= 3(x 2 + 4x + 1) 3 βˆ™(2x + 4 Γ— 1 + 0) + 4(x 3 – 5x – 2) 3 βˆ™(3x 2 – 5 Γ— 1 – 0)
= 6 (x + 2)(x 2 + 4x + 1) 2 + 4 (3x 2 – 5)(x 3 – 5x – 2) 3 .
(ii) (1 + 4x) 5 (3 + x βˆ’ x 2 ) 8
Solution:
Let y = (1 + 4x) 5 (3 + x βˆ’ x 2 ) 8
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-29
= 8 (1 + 4x) 5 (3 + x – x 2 ) 7 βˆ™(0 + 1 – 2x) + 5 (1 + 4x) 4 (3 + x – x 2 ) 8 βˆ™(0 + 4 Γ— 1)
= 8 (1 – 2x)(1 + 4x) 5 (3 + x – x 2 ) 7 + 20(1 + 4x) 4 (3 + x – x 2 ) 8 .

(iii) \(\frac{x}{\sqrt{7-3 x}}\)
Solution:
Let y = \(\frac{x}{\sqrt{7-3 x}}\)
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-30

Maharashtra-Board-Solutions

(iv) \(\frac{\left(x^{3}-5\right)^{5}}{\left(x^{3}+3\right)^{3}}\)
Solution:
Let y = \(\frac{\left(x^{3}-5\right)^{5}}{\left(x^{3}+3\right)^{3}}\)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\left[\frac{\left(x^{3}-5\right)^{5}}{\left(x^{3}+3\right)^{3}}\right]\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-31

(v) (1 + sin 2 x) 2 (1 + cos 2 x) 3
Solution:
Let y = (1 + sin 2 x) 2 (1 + cos 2 x) 3
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-32
= 3(1 + sin 2 x) 2 (1 + cos 2 x) 2 βˆ™[2cosx(-sinx)] + 2 (1 + sin 2 x)(1 + cos 2 x) 3 βˆ™[2sinx-cosx]
= 3 (1 + sin 2 x) 2 (1 + cos 2 x) 2 (-sin 2x) + 2(1 + sin 2 x)(1 + cos 2 x) 3 (sin 2x)
= sin2x (1 + sin 2 x) (1 + cos 2 x) 2 [-3(1 + sin 2 x) + 2(1 + cos 2 x)]
= sin2x (1 + sin 2 x)(1 + cos 2 x) 2 (-3 – 3sin 2 x + 2 + 2cos 2 x)
= sin2x (1 + sin 2 x)(1 + cos 2 x) 2 [-1 – 3 sin 2 x + 2 (1 – sin 2 x)]
= sin 2x(1 + sin 2 x)(1 + cos 2 x) 2 (-1 – 3 sin 2 x + 2 – 2 sin 2 x)
= sin2x (1 + sin 2 x)(1 + cos 2 x) 2 (1 – 5 sin 2 x).

(vi) \(\sqrt{\cos x}+\sqrt{\cos \sqrt{x}}\)
Solution:
Let y = \(\sqrt{\cos x}+\sqrt{\cos \sqrt{x}}\)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}[\sqrt{\cos x}+\sqrt{\cos \sqrt{x}}]\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-33

(vii) log(sec 3x+ tan 3x)
Solution:
Let y = log(sec 3x+ tan 3x)
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-34

(viii) \(\frac{1+\sin x^{\circ}}{1-\sin x^{\circ}}\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-35
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-36

Maharashtra-Board-Solutions

(ix) cot\(\left(\frac{\log x}{2}\right)\) – log\(\left(\frac{\cot x}{2}\right)\)
Solution:
Let y = cot\(\left(\frac{\log x}{2}\right)\) – log\(\left(\frac{\cot x}{2}\right)\)
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-37

(x) \(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-38
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-39

(xi) \(\frac{e^{\sqrt{x}}+1}{e^{\sqrt{x}}-1}\)
Solution:
let y = \(\frac{e^{\sqrt{x}}+1}{e^{\sqrt{x}}-1}\)
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-40

(xii) log[tan 3 xΒ·sin 4 xΒ·(x 2 + 7) 7 ]
Solution:
Let y = log [tan 3 xΒ·sin 4 xΒ·(x 2 + 7) 7 ]
= log tan 3 x + log sin 4 x + log (x 2 + 7) 7
= 3 log tan x + 4 log sin x + 7 log (x 2 + 7)
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-41
= 6cosec2x + 4 cotx + \(\frac{14 x}{x^{2}+7}\)

Maharashtra-Board-Solutions

(xiii) log\(\left(\sqrt{\frac{1-\cos 3 x}{1+\cos 3 x}}\right)\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-42
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-43

(xiv) log\(\left(\sqrt{\left.\frac{1+\cos \left(\frac{5 x}{2}\right)}{1-\cos \left(\frac{5 x}{2}\right)}\right)}\right.\)
Solution:
Using log\(\left(\frac{a}{b}\right)\) = log a – log b
log a b = b log a
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-44
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-45
\(-\frac{5}{2}\)cosec\(\left(\frac{5 x}{2}\right)\)

(xv) log\(\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right)\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-46
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-47
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-48

Maharashtra-Board-Solutions

(xvi) log\(\left[4^{2 x}\left(\frac{x^{2}+5}{\sqrt{2 x^{3}-4}}\right)^{\frac{3}{2}}\right]\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-49
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-50

(xvii) log\(\left[\frac{e^{x^{2}}(5-4 x)^{\frac{3}{2}}}{\sqrt[3]{7-6 x}}\right]\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-51
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-52

(xviii) log\(\left[\frac{a^{\cos x}}{\left(x^{2}-3\right)^{3} \log x}\right]\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-53

(xix) y= (25) log 5 (secx) βˆ’ (16) log 4 (tanx)
Solution:
y = (25) log 5 (secx) βˆ’ (16) log 4 (tanx)
= 5 2log 5 (secx) – 4 2log 4 (tanx)
= 5 log 5 (sec5x) – 4 log 4 (tan2x)
= sec 2 x – tan 2 x … [∡ = x]
∴ y = 1
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(1) = 0

(xx) \(\frac{\left(x^{2}+2\right)^{4}}{\sqrt{x^{2}+5}}\)
Solution:
Let y = \(\frac{\left(x^{2}+2\right)^{4}}{\sqrt{x^{2}+5}}\)
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-54
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-55

Maharashtra-Board-Solutions

Question 4.
A table of values of f, g, f β€˜ and g’ is given
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-56
(i) If r(x) = f [g(x)] find r’ (2).
Solution:
r(x) = f[g(x)]
∴ r'(x) = \(\frac{d}{d x}\)f[g(x)]
= f'[g(x)]βˆ™\(\frac{d}{d x}\)[g(x)]
= f'[g(x)βˆ™[g'(x)]
∴ r'(2) = f'[g(2)]βˆ™g'(2)
= f'(6)βˆ™g'(2) … [∡ g(x) = 6, when x = 2]
= -4 Γ— 4 … [From the table]
= -16.

(ii) If R(x) = g[3 + f(x)] find R’ (4).
Solution:
R(x) = g[3 + f(x)]
∴ R'(x) = \(\frac{d}{d x}\){g[3+f(x)]}
= g'[3 + f(x)]βˆ™\(\frac{d}{d x}\)[3 + f(x)]
= g'[3 +f(x)]βˆ™[0 + f'(x)]
= g'[3 + f(x)]βˆ™f'(x)
∴ R'(4) = g'[3 + f(4)]βˆ™f'(4)
= g'[3 + 3]βˆ™f'(4) … [∡ f(x) = 3, when x = 4]
= g'(6)βˆ™f'(4)
= 7 Γ— 5 … [From the table]
= 35.

Maharashtra-Board-Solutions

(iii) If s(x) = f[9βˆ’ f(x)] find s’ (4).
Solution:
s(x) = f[9βˆ’ f(x)]
∴ s'(x) = \(\frac{d}{d x}\){f[9 – f(x)]}
= f'[9 – f(x)]βˆ™\(\frac{d}{d x}\)[0 – f(x)]
= f'[9 – f(x)]βˆ™[0 – f'(x)]
= -f'[9 – f(x)] – f'(x)
∴ s'(4) = -f'[9 – f(4)] – f'(4)
= -f'[9 – 3] – f'(4) … [∡ f(x) = 3, when x = 4]
= -f'(6) – f'(4)
= -(-4)(5) … [From the table]
= 20.

(iv) If S(x) = g[g(x)] find S’ (6)
Solution:
S(x) = g[g(x)]
∴ S'(x) = \(\frac{d}{d x}\)g[g(x)]
= g'[g(x)]βˆ™\(\frac{d}{d x}\)[g(x)]
= g'[g(x)]βˆ™g'(x)
∴ S β€˜(6) = g'[g'(6)]βˆ™g'(6)
= g'(2)βˆ™g'(6) … [∡ g (x) = 2, when x = 6]
= 4 Γ— 7 … [From the table]
= 28.

Maharashtra-Board-Solutions

Question 5.
Assume that f β€˜(3) = -1, g'(2) = 5, g(2) = 3 and y = f[g(x)] then \(\left[\frac{d y}{d x}\right]_{x=2}\) = ?
Solution:
y = f[g(x)]
∴ \(\frac{d y}{d x}\) = \(\frac{d}{d x}\){[g(x)]}
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-57

Question 6.
If h(x) = \(\sqrt{4 f(x)+3 g(x)}\), f(1) = 4, g(1) = 3, f β€˜(1) = 3, g'(1) = 4 find h'(1).
Solution:
Given f(1) = 4, g(1) = 3, f β€˜(1) = 3, g'(1) = 4 …..(1)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-58

Question 7.
Find the x co-ordinates of all the points on the curve y = sin 2x – 2 sin x, 0 ≀ x < 2Ο€ where \(\frac{d y}{d x}\) = 0.
Solution:
y = sin 2x – 2 sin x, 0 ≀ x < 2Ο€
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-59
= cos2x Γ— 2 – 2cosx
= 2 (2 cos 2 x – 1) – 2 cosx
= 4 cos 2 x – 2 – 2 cos x
= 4 cos 2 x – 2 cos x – 2
If \(\frac{d y}{d x}\) = 0, then 4 cos 2 x – 2 cos x – 2 = 0
∴ 4cos 2 x – 4cosx + 2cosx – 2 = 0
∴ 4 cosx (cosx – 1) + 2 (cosx – 1) = 0
∴ (cosx – 1)(4cosx + 2) = 0
∴ cosx – 1 = 0 or 4cosx + 2 = 0
∴ cos x = 1 or cos x = \(-\frac{1}{2}\)
∴ cos x = cos 0
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.1-60

Maharashtra-Board-Solutions

Question 8.
Select the appropriate hint from the hint basket and fill up the blank spaces in the following paragraph. [Activity]
β€œLet f (x) = x 2 + 5 and g(x) = e x + 3 then
f [g(x)] = _ _ _ _ _ _ _ _ and g [f(x)] =_ _ _ _ _ _ _ _.
Now f β€˜(x) = _ _ _ _ _ _ _ _ and g'(x) = _ _ _ _ _ _ _ _.
The derivative off [g (x)] w. r. t. x in terms of f and g is _ _ _ _ _ _ _ _.
Therefore \(\frac{d}{d x}\)[f[g(x)]] = _ _ _ _ _ _ _ _ _ and [\(\frac{d}{d x}\)[f[g(x)]]] x = 0 = _ _ _ _ _ _ _ _ _ _ _.
The derivative of g[f(x)] w. r. t. x in terms of f and g is _ _ _ _ _ _ __ _ _ _ _.
Therefore \(\frac{d}{d x}\)[g[f(x)]] = _ _ _ _ _ _ _ _ _ and [\(\frac{d}{d x}\)[g[f(x)]]] x = 1 = _ _ _ _ _ _ _ _ _ _ _.”
Hint basket : { f β€˜[g(x)]Β·g'(x), 2e 2x + 6e x , 8, g'[f(x)]Β·f β€˜(x), 2xe x2 + 5 , -2e 6 , e 2x + 6e x + 14, e x2 + 5 + 3, 2x, e x }
Solution:
f[g(x)] = e 2x + 6e x + 14
g[f(x)] = e x2 + 5 + 3
f'(x) = 2x, g’f(x) = e x
The derivative of f[g(x)] w.r.t. x in terms of and g is f'[g(x)]βˆ™g'(x).
∴ \(\frac{d}{d x}\){f[g(x)]} = 2e 2x + 6e x and \(\frac{d}{d x}\){f[g(x)]} x = 0 = 8
The derivative of g[f(x)] w.r.t. x in terms of f and g is g’f(x)]βˆ™f'(x).
∴ \(\frac{d}{d x}\){g[(f(x)]} = 2xe x2 + 5 and
\(\frac{d}{d x}\){g[(f(x)]} x = -1 = -2e 6 .