Higher Engineering Mathematics, Sixth Edition

Methods of differentiation 289

Substituting(x+δx)forxgives
f(x+δx)=(x+δx)^2 =x^2 + 2 xδx+δx^2 , hence

f′(x)=limit

δx→ 0

{

(x^2 + 2 xδx+δx^2 )−(x^2 )

δx

}

=limit

δx→ 0

{

( 2 xδx+δx^2 )

δx

}

=limit

δx→ 0

[2x+δx]

Asδx→0, [2x+δx]→[2x+0]. Thusf′(x)= 2 x,i.e.
the differential coefficient ofx^2 is 2x.Atx=2, the
gradient of the curve,f′(x)= 2 ( 2 )= 4.

Differentiation from first principles can be a lengthy
process and it wouldnot beconvenient togothrough this
procedureeverytimewewant todifferentiateafunction.
In reality we do not have to because a set of general
rules have evolved from the above procedure, which we
consider in the following section.

27.4 Differentiation of common functions

From differentiation by first principles of a number of
examples such as in Problem 1 above, a general rule
for differentiatingy=axnemerges, whereaandnare
constants.

The rule is:ify=axnthen

dy

dx

=anxn−^1

(or,iff(x)=axnthenf′(x)=anxn−^1 ) and is true for all
real values ofaandn.

For example, ify= 4 x^3 thena=4andn=3, and

dy

dx

=anxn−^1 =( 4 )( 3 )x^3 −^1 = 12 x^2

Ify=axnandn=0theny=ax^0 and

dy

dx

=(a)( 0 )x^0 −^1 =0,

i.e.the differential coefficient of a constant is zero.

Figure 27.5(a) shows a graph ofy=sinx. The gra-
dient is continually changing as the curve moves from

0toAtoBtoCtoD. The gradient, given by

dy

dx

,may

be plotted in a corresponding position belowy=sinx,
as shown in Fig. 27.5(b).

y

0

(a)

(b)

0

ysinx

xrad

xrad







A

A

0 

C

B

D

BD

C



2

d

dx



dy

dx

2 

2 

3 

2



 3 

2

(sinx)cosx



2

Figure 27.5

(i) At 0, the gradient is positiveand is at its steepest.

Hence 0′is a maximum positive value.

(ii) Between 0 andAthe gradient is positive but is

decreasing in value until atAthegradient is zero,

shown asA′.

(iii) BetweenAandBthe gradient is negative but

is increasing in value until atBthe gradient is at

its steepest negative value. HenceB′is a maxi-

mum negative value.

(iv) If the gradient ofy=sinxis further investigated

betweenBandDthen the resulting graph of

dy

dx

is seen to be a cosine wave. Hence the rate of

change of sinxis cosx,

i.e.ify=sinxthen

dy

dx

=cosx

By a similar construction to that shown in Fig. 27.5 it

may be shown that:

ify=sinaxthen

dy

dx

=acosax

If graphs ofy=cosx,y=exandy=lnxare plotted

and their gradients investigated, their differential coef-

ficients may be determined in a similar manner to that

shown fory=sinx. The rate of change of a function is

a measure of the derivative.