Higher Engineering Mathematics, Sixth Edition

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.

1. xsinx [xcosx+sinx]


2. x^2 e^2 x [2xe^2 x(x+ 1 )]


3. x^2 lnx [x( 1 +2lnx)]


4. 2x^3 cos3x [6x^2 (cos 3x−xsin3x)]

5.

√

x^3 ln3x

[√

x

(

1 +^32 ln3x

)]

6. e^3 tsin4t [e^3 t(4cos4t+3sin4t)]


7. e^4 θln3θ

[

e^4 θ

(

1

θ

+4ln3θ

)]

8. etlntcost
   [
   et

{(

1

t

+lnt

)

cost−lntsint

}]

9. Evaluate

di

dt

, correct to 4 significant figures,

whent= 0 .1, andi= 15 tsin3t.

[8.732]

10. 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)