Methods of differentiation 293
Hence
dy
dx
=(x^3 cos3x)
(
1
x
)
+(lnx)[− 3 x^3 sin3x
+ 3 x^2 cos3x]
=x^2 cos3x+ 3 x^2 lnx(cos3x−xsin3x)
i.e.
dy
dx
=x^2 {cos3x+3lnx(cos 3x−xsin3x)}
Problem 13. Determine the rate of change of
voltage, givenv= 5 tsin2tvolts whent= 0 .2s.
Rate of change of voltage=
dv
dt
=( 5 t)(2cos2t)+(sin2t)( 5 )
= 10 tcos2t+5sin2t
Whent= 0. 2 ,
dv
dt
= 10 ( 0. 2 )cos2( 0. 2 )+5sin2( 0. 2 )
=2cos0. 4 +5sin0.4(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 whent=0.2s is
3.79volts/s, correct to 3 significant figures.
Now try the following exercise
Exercise 116 Further problemson
differentiating products
In Problems 1 to 8 differentiate the given products
with respect to the variable.
- xsinx [xcosx+sinx]
- x^2 e^2 x [2xe^2 x(x+ 1 )]
- x^2 lnx [x( 1 +2lnx)]
- 2x^3 cos3x [6x^2 (cos 3x−xsin3x)]
5.
√
x^3 ln3x
[√
x
(
1 +^32 ln3x
)]
- e^3 tsin4t [e^3 t(4cos4t+3sin4t)]
- e^4 θln3θ
[
e^4 θ
(
1
θ
+4ln3θ
)]
- etlntcost
[
et
{(
1
t
+lnt
)
cost−lntsint
}]
- Evaluate
di
dt
, correct to 4 significant figures,
whent= 0 .1, andi= 15 tsin3t.
[8.732]
- Evaluate
dz
dt
, correct to 4 significant figures,
whent= 0 .5, given thatz=2e^3 tsin2t.
[32.31]
27.6 Differentiation of a quotient
Wheny=
u
v
,anduandvare both functions ofx
then
dy
dx
=
v
du
dx
−u
dv
dx
v^2
This is known as thequotient rule.
Problem 14. Find the differential coefficient of
y=
4sin5x
5 x^4
4sin5x
5 x^4
is a quotient. Letu=4sin5xandv= 5 x^4
(Note thatvisalwaysthe denominator andu the
numerator.)
dy
dx
=
v
du
dx
−u
dv
dx
v^2
where
du
dx
=( 4 )( 5 )cos 5x=20cos5x
and
dv
dx
=( 5 )( 4 )x^3 = 20 x^3
Hence
dy
dx
=
( 5 x^4 )(20cos5x)−(4sin5x)( 20 x^3 )
( 5 x^4 )^2
=
100 x^4 cos5x− 80 x^3 sin5x
25 x^8
=
20 x^3 [5xcos5x−4sin5x]
25 x^8
i.e.
dy
dx
=
4
5 x^5
(5xcos 5x−4sin5x)