Basic Engineering Mathematics, Fifth Edition

Introduction to differentiation 319

y

(a) 0

(b)

0

y 5 cos x

x radians

x radians

1

2

2



2

1 dydx

 3  2 

2



2

 3  2 

2

dxd (cos x)^52 sin x

Figure 34.6

If each point on the curve y=sinx (as shown in

Figure 34.5(a)) were to be made negative (i.e.+

π

2

made−

π

2

,−

3 π

2

made+

3 π

2

, and so on) then the graph

shown in Figure 34.6(b) would result. This latter graph
therefore represents the curve of−sinx.
Thus,

ify=cosx,

dy

dx

=−sinx

It may also be shown that

ify=cosax,

dy

dx

=−asinax (3)

(whereaisaconstant)

andify=cos(ax+α),

dy

dx

=−asin(ax+α) (4)

(whereaandαare constants).

Problem 16. Find the differential coefficient of

y=7sin2x−3cos4x

dy

dx

=( 7 )(2cos2x)−( 3 )(−4sin4x)

from equations (1) and (3)

=14cos2x+12sin4x

Problem 17. Differentiate the following with

respect to the variable (a)y=2sin5θ

(b)f(t)=3cos2t

(a) y=2sin5θ

dy

dθ

=( 2 )(5cos5θ)=10cos5θ

from equation (1)

(b) f(t)=3cos2t

f′(t)=( 3 )(−2sin2t)=−6sin2t

from equation (3)

Problem 18. Differentiate the following with

respect to the variable

(a)f(θ )=5sin( 100 πθ− 0. 40 )

(b)f(t)=2cos( 5 t+ 0. 20 )

(a) Iff(θ )=5sin( 100 πθ− 0. 40 )

f′(θ)=5[100πcos( 100 πθ− 0. 40 )]

from equation (2), wherea= 100 π

= 500 πcos( 100 πθ− 0. 40 )

(b) Iff(t)=2cos( 5 t+ 0. 20 )

f′(t)=2[−5sin( 5 t+ 0. 20 )]

from equation (4), wherea= 5

=−10sin( 5 t+ 0. 20 )

Problem 19. An alternating voltage is given by

v=100sin200tvolts, wheretis the time in

seconds. Calculate the rate of change of voltage

when (a)t= 0 .005s and (b)t= 0 .01s

v=100sin200tvolts. The rate of change ofvis given

by

dv

dt

dv

dt

=( 100 )(200cos200t)=20000cos200t

(a) Whent= 0 .005s,

dv

dt

=20000cos( 200 )( 0. 005 )=20000cos1.

cos1 means ‘the cosine of 1 radian’ (make sure

yourcalculator is on radians, not degrees). Hence,

dv

dt

=10806 volts per second

(b) Whent= 0 .01s,

dv

dt

=20000cos( 200 )( 0. 01 )=20000cos2.

Hence,

dv

dt

=−8323 volts per second