Methods of differentiation 293
Hence
dy
dx
= (x 3 cos 3x)
1
x
+ (ln x)[−3x 3 sin 3x
+ 3x
2 cos 3x]
= x 2 cos 3x + 3x 2 ln x(cos 3x − x sin 3x)
i.e.
dy
dx
= x 2 {cos 3x + 3 lnx(cos 3x −x sin 3x)}
Problem 13. Determine the rate of change of
voltage, given v = 5t sin 2t volts when t = 0.2 s.
Rate of change of voltage =
dv
dt
= (5t )(2 cos 2t ) + (sin 2t )(5)
= 10t cos 2t + 5 sin2t
When t = 0.2,
dv
dt
= 10(0.2) cos 2(0.2) + 5 sin 2(0.2)
= 2 cos 0.4 + 5 sin 0.4 (wherecos0.4
means the cosine of 0.4 radians)
Hence
dv
dt
= 2(0.92106) + 5(0.38942)
= 1.8421 + 1.9471 = 3.7892
i.e. the rate of change of voltage when t = 0.2 s is
3.79 volts/s, correct to 3 significant figures.
Now try the following exercise
Exercise 116 Further problems on
differentiating products
In Problems 1 to 8 differentiate the given products
with respect to the variable.
1. x sin x
[x cos x + sin x]
2. x 2 e 2x
[2x e 2x (x + 1)]
3. x 2 ln x
[x(1 + 2 ln x)]
4. 2x 3 cos 3x
[6x 2 (cos 3x − x sin 3x)]
5.
√
x 3 ln 3x
√
x
1 +
3
2 ln 3x
6. e 3t sin 4t
[e 3t (4 cos 4t + 3 sin 4t )]
7. e 4θ ln 3θ
e 4θ
1
θ
+ 4 ln3θ
8. e t ln t cos t
e t
1
t
+ ln t
cos t − ln t sin t
9. Evaluate
di
dt
, correct to 4 significant figures,
when t = 0.1, and i = 15t sin 3t .
[8.732]
10. Evaluate
dz
dt
, correct to 4 significant figures,
when t = 0.5, given that z =2e 3t sin 2t .
[32.31]
27.6 Differentiation of a quotient
When y =
u
v
, and u and v are both functions of x
then
dy
dx
=
v
du
dx
− u
dv
dx
v 2
This is known as the quotient rule.
Problem 14. Find the differential coefficient of
y =
4 sin5x
5x 4
4 sin5x
5x 4 is a quotient. Let u = 4 sin5x and v = 5x
4
(Note that v is always the denominator and u the
numerator.)
dy
dx
=
v
du
dx
− u
dv
dx
v 2
where
du
dx
= (4)(5) cos 5x = 20 cos5x
and
dv
dx
= (5)(4)x 3 = 20x 3
Hence
dy
dx
=
(5x 4 )(20 cos 5x) − (4 sin 5x)(20x 3 )
(5x 4 ) 2
=
100x 4 cos 5x − 80x 3 sin 5x
25x 8
=
20x 3 [5x cos 5x − 4 sin 5x]
25x 8
i.e.
dy
dx
=
4
5x 5 (5x cos 5x − 4 sin 5x)