Higher Engineering Mathematics, Sixth Edition

296 Higher Engineering Mathematics

Using the function of a function rule,

dy

dx

=

dy

du

×

du

dx

=

(

1

2

√

u

)

( 6 x+ 4 )=

3 x+ 2

√

u

i.e.

dy

dx

=

3 x+ 2

√

( 3 x^2 + 4 x− 1 )

Problem 22. Differentiatey=3tan^43 x.

Letu=tan3xtheny= 3 u^4

Hence

du

dx

=3sec^23 x,(from Problem 15), and

dy

du

= 12 u^3

Then

dy

dx

=

dy

du

×

du

dx

=( 12 u^3 )(3sec^23 x)

= 12 (tan3x)^3 (3sec^23 x)

i.e.

dy

dx

=36 tan^33 xsec^23 x

Problem 23. Find the differential coefficient of

y=

2

( 2 t^3 − 5 )^4

y=

2

( 2 t^3 − 5 )^4

= 2 ( 2 t^3 − 5 )−^4 .Letu=( 2 t^3 − 5 ),then

y= 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

)

( 6 t^2 )

=

− 48 t^2

(2t^3 −5)^5

Now try the following exercise

Exercise 118 Further problems on the

function of a function

In Problems 1 to 9, find the differential coefficients

with respect to the variable.

1. ( 2 x− 1 )^6 [12( 2 x− 1 )^5 ]


2. ( 2 x^3 − 5 x)^5 [5( 6 x^2 − 5 )( 2 x^3 − 5 x)^4 ]
   3. 2sin( 3 θ− 2 ) [6cos( 3 θ− 2 )]
   4. 2cos^5 α [−10cos^4 αsinα]

5.

1

(x^3 − 2 x+ 1 )^5

[

5 ( 2 − 3 x^2 )

(x^3 − 2 x+ 1 )^6

]

6. 5e^2 t+^1 [10e^2 t+^1 ]


7. 2cot( 5 t^2 + 3 ) [− 20 tcosec^2 ( 5 t^2 + 3 )]


8. 6tan( 3 y+ 1 ) [18sec^2 ( 3 y+ 1 )]


9. 2etanθ [2sec^2 θetanθ]


10. Differentiateθsin

(

θ−

π

3

)

with respect toθ,

and evaluate, correct to 3 significant figures,

whenθ=

π

2

. [1.86]

27.8 Successive differentiation

When a functiony=f(x)is differentiated with respect

toxthedifferentialcoefficient is writtenas

dy

dx

orf′(x).

If the expression is differentiated again, the second dif-

ferential coefficient is obtained and is written as

d^2 y

dx^2

(pronounced dee twoyby deex squared) or f′′(x)

(pronouncedfdouble-dashx).

By successive differentiation further higher derivatives

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)