Methods of differentiation 295
2.
2 cos3x
x 3
−6
x 4 (x sin 3x + cos 3x)
3.
2x
x 2 + 1
2(1 − x 2 )
(x 2 + 1) 2
4.
√
x
cos x
cos x
2
√
x
+
√
x sin x
cos 2 x
5.
3
√
θ 3
2 sin2θ
3
√
θ(3 sin 2θ − 4θ cos 2θ)
4 sin
2 2θ
6.
ln 2t
√
t
⎡
⎢
⎣
1 −
1
2
ln 2t
√
t 3
⎤
⎥
⎦
7.
2xe 4x
sin x
2e 4x
sin
2 x
{(1 + 4x) sin x − x cos x}
8. Find the gradient of the curve y =
2x
x 2 − 5
at
the point (2, −4).
[−18]
9. Evaluate
dy
dx
at x = 2.5, correct to 3 significant
figures, given y =
2x
2
+ 3
ln 2x
.
[3.82]
27.7 Function of a function
It is often easier to make a substitution before differentiating.
If y is a function of x then
dy
dx
=
dy
du
×
du
dx
This is known as the ‘function of a function’ rule (or
sometimes the chain rule).
For example, if y = (3x − 1) 9 then, by making the substitution u = (3x − 1), y = u 9 , which is of the ‘standard’
form.
Hence
dy
du
= 9u 8 and
du
dx
= 3
Then
dy
dx
=
dy
du
×
du
dx
= (9u 8 )(3) = 27u 8
Rewriting u as (3x − 1) gives:
dy
dx
= 27(3x −1)
8
Since y is a function of u, and u is a function of x, then
y is a function of a function of x.
Problem 19. Differentiate y = 3 cos(5x 2 + 2).
Let u =5x 2 + 2 then y = 3 cosu
Hence
du
dx
= 10x and
dy
du
=−3 sinu.
Using the function of a function rule,
dy
dx
=
dy
du
×
du
dx
= (−3 sin u)(10x) = −30x sin u
Rewriting u as 5x 2 + 2 gives:
dy
dx
=−30x sin(5x
2
+ 2)
Problem 20. Find the derivative of
y = (4t 3 − 3t ) 6 .
Let u =4t 3 − 3t , then y = u 6
Hence
du
dt
= 12t 2 − 3 and
dy
du
= 6u 5
Using the function of a function rule,
dy
dx
=
dy
du
×
du
dx
= (6u
5
)(12t
2
− 3)
Rewriting u as (4t 3 − 3t ) gives:
dy
dt
= 6(4t
3
− 3t )
5
(12t
2
− 3)
= 18(4t
2
− 1)(4t
3
− 3t)
5
Problem 21. Determine the differential
coefficient of y =
(3x 2 + 4x − 1).
y =
(3x 2 + 4x − 1) = (3x 2 + 4x − 1)
1
2
Let u =3x 2 + 4x − 1 then y = u
1
2
Hence
du
dx
= 6x + 4 and
dy
du
=
1
2
u
−
1
2 =
1
2
√
u