Higher Engineering Mathematics, Sixth Edition

Methods of differentiation 297

Problem 25. Ify=cosx−sinx,evaluatex,in

the range 0≤x≤

π

2

,when

d^2 y

dx^2

is zero.

Since y=cosx−sinx,

dy

dx

=−sinx−cosx and

d^2 y

dx^2

=−cosx+sinx.

When

d^2 y

dx^2

is zero,−cosx+sinx=0,

i.e. sinx=cosxor

sinx

cosx

=1.

Hence tanx=1andx=arctan1= 45 ◦or

π

4

radsin the

range 0≤x≤

π

2

Problem 26. Giveny= 2 xe−^3 xshow that

d^2 y

dx^2

+ 6

dy

dx

+ 9 y= 0.

y= 2 xe−^3 x(i.e. a product)

Hence

dy

dx

=( 2 x)(−3e−^3 x)+(e−^3 x)( 2 )

=− 6 xe−^3 x+2e−^3 x

d^2 y

dx^2

=[(− 6 x)(−3e−^3 x)+(e−^3 x)(− 6 )]

+(−6e−^3 x)

= 18 xe−^3 x−6e−^3 x−6e−^3 x

i.e.

d^2 y

dx^2

= 18 xe−^3 x−12e−^3 x

Substituting values into

d^2 y

dx^2

+ 6

dy

dx

+ 9 ygives:

( 18 xe−^3 x−12e−^3 x)+ 6 (− 6 xe−^3 x+2e−^3 x)

+ 9 ( 2 xe−^3 x)= 18 xe−^3 x−12e−^3 x− 36 xe−^3 x

+12e−^3 x+ 18 xe−^3 x= 0

Thus wheny= 2 xe−^3 x,

d^2 y

dx^2

+ 6

dy

dx

+ 9 y= 0

Problem 27. Evaluate

d^2 y

dθ^2

whenθ=0given

y=4sec2θ.

Sincey=4sec2θ,

then

dy

dθ

=(4)(2) sec2θtan2θ(from Problem 16)

=8sec2θtan2θ(i.e. a product)

d^2 y

dθ^2

=(8sec2θ)(2sec^22 θ)

+(tan2θ)[( 8 )( 2 )sec 2θtan2θ]

=16sec^32 θ+16sec2θtan^22 θ

When θ=0,

d^2 y

dθ^2

=16sec^30 +16sec0tan^20

= 16 ( 1 )+ 16 ( 1 )( 0 )= 16.

Now try the following exercise

Exercise 119 Further problemson

successive differentiation

1. Ify= 3 x^4 + 2 x^3 − 3 x+2find

(a)

d^2 y

dx^2

(b)

d^3 y

dx^3

.

[(a) 36x^2 + 12 x(b) 72x+12]

2. (a) Given f(t)=

2

5

t^2 −

1

t^3

+

3

t

−

√

t+ 1

determinef′′(t).

(b) Evaluatef′′(t)whent=1.

⎡

⎢

⎣

(a)

4

5

−

12

t^5

+

6

t^3

+

1

4

√

t^3

(b)− 4. 95

⎤

⎥

⎦

In Problems 3 and 4, find the second differential

coefficient with respect to the variable.

3. (a) 3sin2t+cost(b)2ln4θ
   [
   (a)−(12sin2t+cost)(b)

− 2

θ^2

]

4. (a) 2cos^2 x (b)( 2 x− 3 )^4
   [(a) 4(sin^2 x−cos^2 x)(b) 48( 2 x− 3 )^2 ]