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}
]
- Find the gradient of the curvey=
2 x
x^2 − 5
at
the point (2,−4). [−18]
- 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