Higher Engineering Mathematics

288 DIFFERENTIAL CALCULUS

(v) ifFis the point on the curve (1.01,f(1.01)) then

the gradient of chordAF

=

f(1.01)−f(1)

1. 01 − 1

=

1. 0201 − 1

0. 01

= 2. 01

Thus as pointBmoves closer and closer to point
Athe gradient of the chord approaches nearer and
nearer to the value 2. This is called thelimiting value
of the gradient of the chordABand whenBcoin-
cides withAthe chord becomes the tangent to the
curve.

27.2 Differentiation from first

principles

In Fig. 27.4,AandBare two points very close

together on a curve,δx(deltax) andδy(deltay) rep-

resenting small increments in thexandydirections,

respectively.

Figure 27.4

Gradient of chordAB=

δy

δx

; however,

δy=f(x+δx)−f(x).

Hence

δy

δx

=

f(x+δx)−f(x)

δx

.

Asδxapproaches zero,

δy

δx

approaches a limiting

value and the gradient of the chord approaches the

gradient of the tangent atA.

When determining the gradient of a tangent to a

curve there are two notations used. The gradient of

the curve atAin Fig. 27.4 can either be written as

limit

δx→ 0

δy

δx

or limit

δx→ 0

{

f(x+δx)−f(x)

δx

}

InLeibniz notation,

dy

dx

=limit

δx→ 0

δy

δx

Infunctional notation,

f′(x)=limit

δx→ 0

{

f(x+δx)−f(x)

δx

}

dy

dx

is the same asf′(x) and is called thedifferential

coefficientor thederivative. The process of finding

the differential coefficient is calleddifferentiation.

Problem 1. Differentiate from first principle

f(x)=x^2 and determine the value of the gradient

of the curve atx=2.

To ‘differentiate from first principles’ means ‘to find

f′(x)’ by using the expression

f′(x)=limit

δx→ 0

{

f(x+δx)−f(x)

δx

}

f(x)=x^2

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

{

(2xδx+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.

27.3 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, whereaand

nare constants.

The rule is:ify=axnthen

dy

dx

=anxn−^1