Higher Engineering Mathematics

296 DIFFERENTIAL CALCULUS

y=

2

(2t^3 −5)^4

=2(2t^3 −5)−^4. Let u=(2t^3 −5),

theny= 2 u−^4

Hence

du

dt

= 6 t^2 and

dy

du

=− 8 u−^5 =

− 8

u^5

Then

dy

dt

=

dy

du

×

du

dt

=

(

− 8

u^5

)

(6t^2 )

=

− 48 t^2

(2t^3 −5)^5

Now try the following exercise.

Exercise 120 Further problems on the func-

tion of a function

In Problems 1 to 8, find the differential coeffi-

cients with respect to the variable.

1. (2x^3 − 5 x)^5 [5(6x^2 −5)(2x^3 − 5 x)^4 ]


2. 2 sin (3θ−2) [6 cos (3θ−2)]


3. 2 cos^5 α [−10 cos^4 αsinα]

4.

1

(x^3 − 2 x+1)^5

[

5(2− 3 x^2 )

(x^3 − 2 x+1)^6

]

5. 5e^2 t+^1 [10e^2 t+^1 ]


6. 2 cot (5t^2 +3) [− 20 tcosec^2 (5t^2 +3)]


7. 6 tan (3y+1) [18 sec^2 (3y+1)]


8. 2etanθ [2 sec^2 θetanθ]


9. Differentiateθsin

(

θ−

π

3

)

with respect toθ,

and evaluate, correct to 3 significant figures,

whenθ=

π

2

[1.86]

27.7 Successive differentiation

When a function y=f(x) is differentiated with
respect toxthe differential coefficient is written as
dy
dx

orf′(x). If the expression is differentiated again,

the second differential coefficient is obtained and

is written as

d^2 y

dx^2

(pronounced dee twoyby deex

squared) orf′′(x) (pronouncedfdouble-dashx).

By successive differentiation further higher deriv-

atives such as

d^3 y

dx^3

and

d^4 y

dx^4

may be obtained.

Thus ify= 3 x^4 ,

dy

dx

= 12 x^3 ,

d^2 y

dx^2

= 36 x^2 ,

d^3 y

dx^3

= 72 x,

d^4 y

dx^4

=72 and

d^5 y

dx^5

=0.

Problem 24. Iff(x)= 2 x^5 − 4 x^3 + 3 x−5, find

f′′(x).

f(x)= 2 x^5 − 4 x^3 + 3 x− 5

f′(x)= 10 x^4 − 12 x^2 + 3

f′′(x)= 40 x^3 − 24 x= 4 x(10x^2 −6)

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=1 andx=arctan1= 45 ◦ or

π

4

rads

in 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

=(2x)(−3e−^3 x)+(e−^3 x)(2)

=− 6 xe−^3 x+2e−^3 x