Higher Engineering Mathematics, Sixth Edition

Methods of differentiation 295

2.

2cos3x

x^3

[

− 6

x^4

(xsin3x+cos3x)

]

3.

2 x

x^2 + 1

[

2 ( 1 −x^2 )

(x^2 + 1 )^2

]

4.

√

x

cosx

[cosx

2

√

x +

√

xsinx

cos^2 x

]

5.

3

√

θ^3

2sin2θ

[

3

√

θ(3sin2θ− 4 θcos2θ)

4sin^22 θ

]

6.

ln2t

√

t

⎡

⎢

⎣

1 −

1

2

ln2t

√

t^3

⎤

⎥

⎦

7.

2 xe^4 x

sinx

[

2e^4 x

sin^2 x

{( 1 + 4 x)sinx−xcosx}

]

8. Find the gradient of the curvey=

2 x

x^2 − 5

at

the point (2,−4). [−18]

9. Evaluate

dy

dx

atx= 2 .5, correct to 3 significant

figures, giveny=

2 x^2 + 3

ln2x

[3.82]

27.7 Function of a function

It is often easier to make a substitution before differen-
tiating.

Ifyis a function ofxthen

dy

dx

=

dy

du

×

du

dx

This is known as the‘function of a function’rule (or
sometimes thechain rule).
For example, ify=( 3 x− 1 )^9 then,bymakingthesub-
stitutionu=( 3 x− 1 ),y=u^9 , which is of the ‘standard’
form.

Hence

dy

du

= 9 u^8 and

du

dx

= 3

Then

dy

dx

=

dy

du

×

du

dx

=( 9 u^8 )( 3 )= 27 u^8

Rewritinguas( 3 x− 1 )gives:

dy

dx

=27(3x−1)^8

Sinceyis a function ofu,anduis a function ofx,then

yis a function of a function ofx.

Problem 19. Differentiatey=3cos( 5 x^2 + 2 ).

Letu= 5 x^2 +2theny=3cosu

Hence

du

dx

= 10 xand

dy

du

=−3sinu.

Using the function of a function rule,

dy

dx

=

dy

du

×

du

dx

=(−3sinu)( 10 x)=− 30 xsinu

Rewritinguas 5x^2 +2gives:

dy

dx

=− 30 xsin( 5 x^2 + 2 )

Problem 20. Find the derivative of

y=( 4 t^3 − 3 t)^6.

Letu= 4 t^3 − 3 t,theny=u^6

Hence

du

dt

= 12 t^2 −3and

dy

du

= 6 u^5

Using the function of a function rule,

dy

dx

=

dy

du

×

du

dx

=( 6 u^5 )( 12 t^2 − 3 )

Rewritinguas( 4 t^3 − 3 t)gives:

dy

dt

= 6 ( 4 t^3 − 3 t)^5 ( 12 t^2 − 3 )

=18(4t^2 −1)(4t^3 − 3 t)^5

Problem 21. Determine the differential

coefficient ofy=

√

( 3 x^2 + 4 x− 1 ).

y=

√

( 3 x^2 + 4 x− 1 )=( 3 x^2 + 4 x− 1 )

1

2

Letu= 3 x^2 + 4 x−1theny=u

1

2

Hence

du

dx

= 6 x+4and

dy

du

=

1

2

u−

1

(^2) =
1
2
√
u