Question 1:

Find the second order derivatives of each of the following functions:

(i) x3 + tan x
(ii) sin (log x)
(iii) log (sin x)
(iv) ex sin 5x
(v) e6x cos 3x
(vi) x3 log x
(vii) tan−1 x
(viii) x cos x
(ix) log (log x)

Answer 1:
(i) We have,

y = x3 + tan x
Differentiating w. r. t. x, we get
dy
dx
= 3x2 + sec2 x
Differentiating again w. r. t. x, we get
d2 y
d x2
= 6x + 2 sec2 x tan x

(ii) We have,

y = sin (log x)
Differentiating w. r. t. x, we get
dy 1
dx
= cos (log x) × x
Differentiating again w. r. t. x, we get
d2 y −1
d x2
= − sin (log x) 1x × 1
x
+ cos (log x) × x2
−[sin(log x)+cos(log x)]
= x2

(iii) We have,

y = log (sin x)
Differentiating w. r. t. x, we get
dy 1
dx
= sin x
× cos x = cot x
Differentiating again w. r. t. x, we get
d2 y
d x2
= − cosec2 x

(iv) We have,
 y = ex sin (5x)
Differentiating w. r. t. x, we get
dy
dx
= ex sin 5x + ex cos 5x × 5
Differentiating again w. r. t. x, we get
d2 y
d x2
= ex sin 5x + ex cos 5x × 5 + 5ex (− sin 5x × 5) + 5ex cos 5x
= −24ex sin 5x + 10ex cos 5x
= 2ex (5 cos 5x − 12 sin 5x)

(v) We have,

y = e6x cos 3x
Differentiating w. r. t. x, we get
dy
dx
= e6x × 6 × cos 3x + e6x (− sin 3x × 3)
= 6e6x cos 3x − 3e6x sin 3x
Differentiating again w. r. t. x, we get
d2 y
d x2
= 6e6x cos 3x × 6 − 6e6x sin 3x × 3 − 3 × 6 e6x sin 3x − 3e6x × 3 cos 3x
= 27e6x cos 3x − 36e6x sin 3x
= 9e6x (3 cos 3x − 4 sin 3x)

(vi) We have,
y = x3 log x
Differentiating w. r. t. x, we get
dy 1
dx
= 3x2 log x + x3 × x
= 3x2 log x + x2
Differentiating again w. r. t. x, we get
d2 y 1
d x2
= 6x log x + 3x2 × x
+ 2x
= 6x log x + 5x

(vii) We have,
y = tan−1 x
Differentiating w. r. t. x, we get
dy 1
dx
= 1+x2
Differentiating again w. r. t. x, we get
d2 y 1 −2x
d x2
= − (2x) × =
(1+x2 )2 (1+x2 )2

(viii) We have,
y = x cos x
Differentiating w. r. t. x, we get
dy
dx
= cos x − x sin x
Differentiating again w. r. t. x, we get
d2 y
d x2
= − sin x − sin x − x cos x
= − (2 sin x + x cos x)
 (ix) We have,

y = log (log x)
Differentiating w. r. t. x, we get
dy 1 1 1
dx
= log x
× x
= x log x
Differentiating again w. r. t. x, we get
d2 y 0−(log x+1) (1+log x)
d x2
= 2 =− 2
(x log x) (x log x)

Question 2:
2
−x d y −x
If y = e cos x, show that 2
= 2e sin x.
dx

Answer 2:
Here,

y = e−x cos x
Differentiating w. r. t. x, we get
dy
dx
= −e−x sin x − e−x cos x
= − (e−x sin x + e−x cos x)
Differentiating again w. r. t. x, we get
d2 y
d x2
= − (e−x cos x − e−x sin x − e−x sin x − e−x cos x)
= 2e−x sin x

Hence proved.

Question 3:
2
2 d y
If y = x + tan x, show that cos x 2
− 2y + 2x = 0.
dx

Answer 3:
Here,

y = x + tan x
Differentiating w. r. t. x, we get
dy
dx
= 1 + sec2 x
Differentiating again w. r. t. x, we get
d2 y
d x2
= 2 sec2 x tan x
Dividing both sides by sec2 x, we get
d2 y
cos2 x d x2
= 2 tan x
d2 y
⇒ cos2 x d x2
= 2(y − x) [∵ y = x + tan x ⇒ tan x = y − x]
d2 y
⇒ cos2 x d x2
− 2y + 2x = 0
 Hence proved.

Question 4:
4
3 d y 6
If y = x log x, prove that 4
= x
.
dx

Answer 4:
Here,
y = x3 log x
Differentiating w. r. t. x, we get
dy 1
dx
= 3x2 log x + x3 × x
= 3x2 log x + x2
Differentiating again w. r. t. x, we get
d2 y 1
d x2
= 6x log x + 3x2 × x
+ 2x
= 6x log x + 5x
Differentiating again w. r. t. x, we get
d3 y 1
d x3
= 6 log x + 6x × x
+ 5 = 6 log x + 11
Differentiating again w. r. t. x, we get
d4 y 6
d x4
= x

Hence proved.

Question 5:
3
d y 3
If y = log (sin x), prove that 3
= 2 cos x cosec x.
dx

Answer 5:
Here,
y = log (sin x)
Differentiating w. r. t. x, we get
dy 1
dx
= sin x
× cos x = cot x
Differentiating again w. r. t. x, we get
d2 y
d x2
= − cosec2 x
Differentiating again w. r. t. x, we get
d3 y
d x3
= −2 cosec x × (− cosec x cot x)
= 2 cot x cosec2 x = 2 cos x cosec3 x

Hence proved.

Question 6:
2
d y
If y = 2 sin x + 3 cos x, show that 2 + y = 0.
dx
Answer 6:
Here,
y = 2 sin x + 3 cos x
Differentiating w. r. t. x, we get
dy
dx
= 2 cos x − 3 sin x
Differentiating again w. r. t. x, we get
d2 y
d x2
= −2 sinx − 3 cosx
d2 y
⇒ d x2
= − ( 2 sin x + 3 cos x)
d2 y
⇒ d x2
= −y
d2 y
⇒ d x2
+y=0

Hence proved.

Question 7:
2
log x d y 2 log x−3
If y = x
, show that 2
= 3
.
dx x

Answer 7:
Here,

log x
y= x
Differentiating w. r. t. x, we get
dy 1−log x
dx
= x2
Differentiating again w. r. t. x, we get
d2 y −x−2x(1−log x)
d x2
= x4
−x−2x+2x log x
= x4
−3+2 log x
= x3
2 log x−3
= x3

Hence proved.

Question 8:
2 4
d y b
If x = a sec θ, y = b tan θ, prove that 2
=− 2 3
.
dx a y

Answer 8:
Here,
 x = a sec θ and y = b tan θ
Differentiating w. r. t. θ, we get
dx dy
dθ
= a sec θ tan θ and dθ
= b sec2 θ
dy dy dθ b sec2 θ b cosec θ
∴ dx
= dθ
× dx
= a sec θ tan θ
= a
Differentiating w. r. t. x, we get
d2 y b dθ
d x2
= a
× − cosec θ cot θ × dx
1
= − ba × cosec θ cot θ × a sec θ tan θ
= − ab2 × cot θ × 1
tan2 θ
1
= − ab2 × tan3 θ
−b4
= a2 y 3
[∵ y = b tan θ]

Hence proved.

Question 9:
If x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ), prove that
d2x d2y d2y sec3 θ
dθ2
= a (cos θ − θ sin θ), dθ2
= a (sin θ + θ cos θ) and dx2
= aθ
.

Answer 9:
We have, x = a (cos θ + θ sin θ)
dx d[a(cos θ+θ sin θ)]
dθ
= dθ
dx
⇒ dθ
= −a sin θ + a sin θ + aθ cos θ = aθ cos θ . . . . . (i)
d2x d dx d(aθ cos θ)
dθ2
= dθ
( dθ )= dθ
d2x
⇒ dθ2
= a cos θ − aθ sin θ = a (cos θ − θ sin θ)

y = a (sin θ − θ cos θ)
dy d[a(sin θ−θ cos θ)]
dθ
= dθ
dy
⇒ dθ
= a cos θ − a cos θ + aθ sin θ = aθ sin θ . . . . . (ii)
d2y d dy d(aθ sin θ)
dθ2
= dθ
( dθ ) = dθ
d2y
⇒ dθ2
= a sin θ + aθ cos θ = a (sin θ + θ cos θ)

From (i) and (ii), we have

dy
dy /dθ
aθ sin θ
dx
= dx
= aθ cos θ
= tan θ
/dθ

d2y d dy d(tan θ)
2 = dx
( dx ) = dx
= sec2 θ dθ
dx
dx
d2y 1
⇒ 2
= (sec2 θ) ( aθ cos θ
)
dx
d2y sec3 θ
⇒ = aθ
dx2

Hence proved.

Question 10:
 2
d y x π
x
If y = e cos x, prove that 2
= 2 e cos (x + 2
).
dx

Answer 10:
Here,
y = ex cos x
Differentiating w. r. t. x, we get
dy
dx
= ex cos x − ex sin x = ex (cos x − sin x)
Differentiating again w. r. t. x, we get
d2 y
d x2
= ex (cos x − sin x) + ex (− sin x − cos x)
= ex cos x − ex sin x − ex sin x − ex cos x
= −2ex sin x
= 2ex cos (x + π2 )

Hence proved.

Question 11:
2 4
d y b
If x = a cos θ, y = b sin θ, show that 2
=− 2 3
.
dx a y

Answer 11:
Here,

x = a cos θ and y = b sin θ

Differentiating w. r. t. θ, we get
dx dy
dθ
= − a sin θ and dθ
= b cos θ
dy b cos θ −b
∴ dx
= −a sin θ
= a cot θ
Differentiating w. r. t. x, we get
d2 y
d x2
= − ba × (− cosec2 θ) dθ
dx
b 1
= a
× cosec2 θ × −a sin θ

= − ab2 × 1
sin3 θ
b3
= − ab2 × y3
[∵ y = b sin θ]
−b4
= a2 y 3

Hence proved.

Question 12:
2
π
at θ =
3 3 d y 32
If x = a (1 − cos θ), y = a sin θ, prove that 2 = 27a 6
.
dx

Answer 12:
Here,
 x = a (1 − cos3 θ), y = a sin3 θ
Differentiating w. r. t. θ, we get
dx dy
dθ
= 3a cos2 θ sin θ and dθ
= 3a sin2 θ cos θ
dy 3a sin2 θ cos θ
⇒ dx
= 3a cos2 θ sin θ
= tan θ
Differentiating w. r. t. x, we get
d2 y dθ
d x2
= sec2 θ dx
sec2 θ
= 3a cos2 θ sin θ
sec4 θ
= 3a sin θ
d2 y π
∴ d x2
at θ = 6
4
π
2 (sec )
d y 6 32
⇒ d x2
= 3a sin π = 27a
6

Question 13:
2
d y a
If x = a (θ + sin θ), y = a (1 + cos θ), prove that 2 =− 2 .
dx y

Answer 13:
Here,
x = a (θ + sin θ) and y = a (1 + cos θ)
Differentiating w. r. t. θ, we get
dx dy
dθ
= a + a cos θ and dθ
= −a sin θ
dy −a sin θ − sin θ
∴ dx
= a+a cos θ
= 1+cos θ
Differentiating w. r. t. x, we get
d2 y (1+cos θ) cos θ+ sin2 θ
d x2
= −{ 2 } dθ
dx
(1+cosθ)
− cos θ−cos2 θ− sin2 θ 1
= × a+a cos θ
(1+cos θ)2
−(1+cos θ)
= 3
a(1+cos θ)
−1
= 2
a(1+cos θ)
−a
= y2
[∵ y = a (1 + cos θ)]

Hence proved.

Question 14:
2
d y
If x = a (θ − sin θ), y = a (1 + cos θ) prove that, find 2 .
dx

Answer 14:
Here,
 x = a (θ − sin θ) and y = a (1 + cos θ)
Differentiating w. r. t. θ, we get
dx dy
dθ
= a − a cos θ, dθ
= −a sin θ
dy −a sin θ − sin θ
⇒ dx
= a−a cos θ
= 1−cos θ
Differentiating w. r. t. x, we get
d2 y − cos θ+cos2 θ+ sin2 θ dθ
d x2
= 2 × dx
(1−cos θ)
− cos θ+cos2 θ+ sin2 θ 1
= × a−a cos θ
(1−cos θ)2
(1−cos θ)
= 3
a(1−cosθ)
1
= 2
a(1−cos θ)

Question 15:
2
π
= − a at θ =
d y 1
If x = a(1 − cos θ), y = a(θ + sin θ), prove that 2 2
.
dx

Answer 15:
Here,
x = a (1 − cos θ) and y = a (θ + sin θ)
Differentiating w. r. t. x, we get
dx dy
dθ
= a sin θ and dθ
=a+ a cos θ
dy a+a cos θ 1+cos θ
∴ dx
= a sin θ
= sin θ
Differentiating w. r. t. x, we get
d2 y − sin2 θ−(1+cos θ) cos θ
d x2
={ 2 } dd θx
sin θ
(1+cosθ)cosθ 1
= (−1 − 2 ) a sin θ
sin θ
d2 y π
∴ d x2
at θ = 2
d2 y −1
⇒[ ]
d x2 θ= π
= 1
a
(−1 − 01 ) = a
2

Hence proved.