METHODS OF DIFFERENTIATION 295

G

27.6 Function of a function

It is often easier to make a substitution before
differentiating.

If y is 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=(3x−1)^9 then, by making

the substitutionu=(3x−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

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

Rewritinguas (3x−1) gives:

dy

dx

=27(3x−1)^8

Sinceyis a function ofu, anduis a function ofx,
thenyis a function of a function ofx.

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

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

Hence

du

dx

= 10 xand

dy

du

=−3 sinu.

Using the function of a function rule,

dy

dx

=

dy

du

×

du

dx

=(−3 sinu)(10x)=− 30 xsinu

Rewritinguas 5x^2 +2 gives:

dy

dx

=− 30 xsin( 5 x^2 + 2 )

Problem 20. Find the derivative of

y=(4t^3 − 3 t)^6.

Letu= 4 t^3 − 3 t, theny=u^6

Hence

du

dt

= 12 t^2 −3 and

dy

du

= 6 u^5

Using the function of a function rule,

dy

dx

=

dy

du

×

du

dx

=(6u^5 )(12t^2 −3)

Rewritinguas (4t^3 − 3 t) gives:

dy

dt

=6(4t^3 − 3 t)^5 (12t^2 −3)

=18(4t^2 −1)(4t^3 − 3 t)^5

Problem 21. Determine the differential coeffi-

cient ofy=

√

(3x^2 + 4 x−1).

y=

√

(3x^2 + 4 x−1)=(3x^2 + 4 x−1)

1

2

Letu= 3 x^2 + 4 x−1 theny=u

1

2

Hence

du

dx

= 6 x+4 and

dy

du

=

1

2

u−

1

(^2) =
1
2
√
u
Using the function of a function rule,
dy
dx

dy
du
×
du
dx

(
1
2
√
u
)
(6x+4)=
3 x+ 2
√
u
i.e.
dy
dx

3 x+ 2
√
( 3 x^2 + 4 x− 1 )
Problem 22. Differentiatey=3 tan^43 x.
Letu=tan 3xtheny= 3 u^4
Hence
du
dx
=3 sec^23 x, (from Problem 15), and
dy
du
= 12 u^3
Then
dy
dx

dy
du
×
du
dx
=(12u^3 )(3 sec^23 x)
=12( tan 3x)^3 (3 sec^23 x)
i.e.
dy
dx
=36 tan^33 xsec^23 x
Problem 23. Find the differential coefficient of
y=
2
(2t^3 −5)^4