Basic Engineering Mathematics, Fifth Edition

320 Basic Engineering Mathematics

Now try the following Practice Exercise

PracticeExercise 134 Differentiation of

sine and cosine functions (answers on

page 354)

1. Differentiatewithrespect tox:(a)y=4sin3x
   (b)y=2cos6x.


2. Givenf(θ )=2sin3θ−5cos2θ,findf′(θ ).


3. Find the gradient of the curvey=2cos

1

2

xat

x=

π

2

4. Determine the gradient of the curve

y=3sin2xatx=

π

3

5. An alternating current is given by
   i=5sin100tamperes, wheretis the time
   in seconds. Determine the rate of change of
   current

(

i.e.

di

dt

)

whent= 0 .01 seconds.

6. v=50sin40tvolts represents an alternating
   voltage,v,wheretis the time in seconds. At
   a time of 20× 10 −^3 seconds, find the rate of
   change of voltage

(

i.e.

dv

dt

)

7. If f(t)=3sin( 4 t+ 0. 12 )−2cos( 3 t− 0. 72 ),
   determinef′(t).

34.7 Differentiation ofeaxand lnax

A graph ofy=exis shown in Figure 34.7(a). The gra-

dient of the curve at any point is given by

dy

dx

and is

continually changing. By drawing tangents to the curve

at many points on the curve and measuring the gradient

ofthetangents,valuesof

dy

dx

forcorrespondingvaluesof

xmay be obtained. These values are shown graphically

in Figure 34.7(b).

The graph of

dy

dx

againstxis identical to the original

graph ofy=ex. It follows that

ify=ex,then

dy

dx

=ex

It may also be shown that

ify=eax,then

dy

dx

=aeax

3

y 5 ex

1 2 x

5

10

15

20

y

23022 21

(a)

3

dy

dx^5 e

x

dy

dx

1 2 x

5

10

15

20

y

23022 21

(b)

Figure 34.7

Therefore,

ify= 2 e^6 x,then

dy

dx

=( 2 )( 6 e^6 x)= 12 e^6 x

A graph ofy=lnxis shown in Figure 34.8(a). The gra-

dient of the curve at any point is given by

dy

dx

and is

continually changing. By drawing tangents to the curve

at many points on the curve and measuring the gradient

of the tangents, values of

dy

dx

for corresponding values

ofxmay be obtained. These values are shown graphi-

cally in Figure 34.8(b).

The graph of

dy

dx

againstxis the graph of

dy

dx

=

1

x

It follows that

ify=lnx,then

dy

dx

=

1

x

It may also be shown that

ify=lnax,then

dy

dx

=

1

x

(Note that, in the latter expression, the constantadoes

not appear in the

dy

dx

term.) Thus,

ify=ln4x,then

dy

dx

=

1

x