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Home Work 01

This document provides examples and solutions for differentiating various composite functions. It defines several standard derivatives of composite functions and then works through multiple examples applying the rules to find derivatives of more complex expressions involving combinations of trigonometric, exponential, logarithmic and power functions.

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JEE Mathematics
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11 views7 pages

Home Work 01

This document provides examples and solutions for differentiating various composite functions. It defines several standard derivatives of composite functions and then works through multiple examples applying the rules to find derivatives of more complex expressions involving combinations of trigonometric, exponential, logarithmic and power functions.

Uploaded by

JEE Mathematics
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
You are on page 1/ 7

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1.1.3 Derivatives of some standard Composite Functions :

dy dy
y y
dx dx
[ f (x)] n n [ f (x)] n−1 ⋅ f '(x) cot [ f (x)] − cosec2 [ f (x)]⋅ f '(x)
f '(x)
cosec [ f (x)] − cosec [ f (x)] ⋅ cot [ f (x)] ⋅ f '(x)
√ f (x) 2√ f (x)
a f (x) a f (x) ⋅ log a ⋅ f '(x)
1 n ⋅ f '(x)
− e f (x) e f (x) ⋅ f '(x)
[ f (x)]n [ f (x)] n+1
f '(x)
sin [ f (x)] cos [ f (x)]⋅ f '(x) log [ f (x)]
cos [ f (x)] − sin [ f (x)]⋅ f '(x) f (x)
tan [ f (x)] sec2 [ f (x)]⋅ f '(x) f '(x)
log a [ f (x)]
sec [ f (x)] sec [ f (x)] ⋅ tan [ f (x)] ⋅ f '(x) f (x) log a
Table 1.1.2
SOLVED EXAMPLES

Ex. 1 : Differentiate the following w. r. t. x.


(i) y = √ x2 + 5 (ii) y = sin (log x) (iii) y = e tan x
3
(iv) log (x5 + 4) (v) 53 cos x − 2 (vi) y=
(2x2 − 7)5
Solution : (i) y = √ x2 + 5
Method 1 : Method 2 :
Let u = x2 + 5 then y = √ u , where y is We have y = √ x2 + 5
a differentiable function of u and u is a Differentiate w. r. t. x
differentiable function of x then
. . . . . (I)
[Treat x2 + 5 as u in mind and use the formula
Now, y = √ u
of derivative of √ u ]
Differentiate w. r. t. u

and u = x2 + 5

Differentiate w. r. t. x
du d 2
= (x + 5) = 2x
dx dx
Now, equation (I) becomes,

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(ii) y = sin (log x)


Method 1 : Method 2 :
Let u = log x then y = sin u, where y is We have y = sin (log x)
a differentiable function of u and u is a Differentiate w. r. t. x
differentiable function of x then
[sin (log x)]
. . . . . (I)
[Treat log x as u in mind and use the formula
Now, y = sin u of derivative of sin u]
Differentiate w. r. t. u

= cos u and u = log x


Differentiate w. r. t. x
1
=
x
Now, equation (I) becomes,

Note : Hence onwards let's use Method 2.


(iii) y = e tan x (iv) Let y = log (x5 + 4)
Differentiate w. r. t. x Differentiate w. r. t. x
[e tan x] [log (x5 + 4)]

3
(v) Let y = 53 cos x − 2 (vi) Let y =
(2x2 − 7)5
Differentiate w. r. t. x
Differentiate w. r. t. x
[53 cos x − 2]
dy
= 53 cos x − 2 · log 5 × (3 cos x − 2)
dx
dy

= − 3 sin x · 53 cos x − 2 · log 5
dx

5

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https://www.ncertbooks.guru/

Ex. 2 : Differentiate the following w. r. t. x.


(i) y = √ sin x3 (ii) y = cot2 (x3) (iii) y = log [cos (x5)]
(iv) y = (x3 + 2x − 3)4 (x + cos x) 3 (v) y = (1 + cos2 x) 4 × √ x + √tan x
Solution :

(i) y= √ sin x3 (ii) y = cot2 (x3)


Differentiate w. r. t. x Differentiate w. r. t. x



= 2 cot (x3) [cot (x3)]

= 2 cot (x3)[− cosec2 (x3)] (x3)



= − 2 cot (x3)cosec2 (x3)(3x2)

dy
∴ ∴ = − 6x2 cot (x3)cosec2 (x3)
dx

(iii) y = log [cos (x5)]

Differentiate w. r. t. x
dy
= (log [cos (x5)])
dx


dy
∴ = − tan (x5) (5x4) = − 5x4 tan (x5)
dx

(iv) y = (x3 + 2x − 3)4 (x + cos x) 3

Differentiate w. r. t. x

= (x3 + 2x − 3)4 ⋅ (x + cos x) 3 + (x + cos x) 3⋅ (x3 + 2x − 3)4

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https://www.ncertbooks.guru/

= (x3 + 2x − 3)4 ⋅3 (x + cos x) 2 ⋅ (x + cos x) + (x + cos x) 3⋅ 4(x3 + 2x − 3)3 ⋅ (x3 + 2x − 3)

= (x3 + 2x − 3)4 ⋅3 (x + cos x) 2 (1 − sin x) + (x + cos x) 3 ⋅ 4(x3 + 2x − 3)3 (3x2 + 2)


dy
∴ = 3(x3 + 2x − 3)4 (x + cos x) 2 (1 − sin x) + 4 (3x2 + 2) (x3 + 2x − 3)3 (x + cos x)3
dx
(v) y = (1 + cos2 x) 4 × √ x + √tan x
Differentiate w. r. t. x







Ex. 3 : Differentiate the following w. r. t. x.

(i) y = log3 (log5 x) (ii)

(iii) (iv)

(v) (vi)

7

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https://www.ncertbooks.guru/

Solution :
(i) y = log3 (log5 x)

= log3 = log3 (log x) - log3 (log 5)

∴ y = - log3 (log 5)
Differentiate w. r. t. x



[Note that log3(log 5) is constant]


∴

(ii)


∴ [ ⸪ log e = 1]

Differentiate w. r. t. x




∴

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https://www.ncertbooks.guru/

(iii)

∴
Differentiate w. r. t. x





∴

(iv)



∴
Differentiate w. r. t. x

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(v) (vi)
y = a cot x [⸪ alog a f (x) = f (x)]

Differentiate w. r. t. x

[⸪ alog a f (x) = f (x)] (a cot x )

= sin2 x + cos2 x = a cot x log a · (cot x)
∴ y =1
Differentiate w. r. t. x =a cot x log a (− cosec2 x)

− cosec2 x· a cot x log a

Ex. 4 : If f (x) = √ 7g (x) − 3 , g (3) = 4 and g' (3) = 5, find f ' (3).

Solution : Given that : f (x) = √ 7g (x) − 3


Differentiate w. r. t. x

∴

For x = 3, we get
35 7
= = [Since g (3) = 4 and g' (3) = 5]
2(5) 2

10

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