THE DIFFERENTIALS

Consider a function defined by y=f(x) where x is the independent variable. In the four-step rule we introduced the symbol Δx to the denote the increment of x. Now we introduce the symbol dx which we call the differential of x. Similarly, we shall call the symbol dy as the differential of y. To give separate meanings to dx and dy, we shall adopt the following definitions of a function defined by the equation y=f(x).

DEFINITION 1: dx = Δx In words, the differential of the independent variable is equal to the increment of the variable.

DEFINITION 2: dy = f’ (x) dx In words, the differential of a function is equal to its derivative multiplied by the differential of its independent variable.

We emphasize that the differential dx is also an independent variable, it may be assigned any value whatsoever. Therefore, from DEFINITION 2, we see that the differential dy is a function of two independent variables x and dx. It should also be noted that while dx=Δx, dy≠Δy in general. Suppose dx≠0 and we divide both sides of the equation dy = f’ (x) dx

by dx. Then we get( )x'f

dxdy =

Note that this time dy/dx denotes the quotient of two differentials, dy and dx . Thus the definition of the differential makes it possible to define the derivative of the function as the ratio of two differentials. That is,

( )xof aldifferenti theyof aldifferenti the

dxdy

x'f ==

The differential may be given a geometric interpretation. Consider again the equation y=f(x) and let its graph be as shown below. Let P(x,y) and Q(x+Δx,f(x)+Δx) be two points on the curve. Draw the

tangent to the curve at P. Through Q, draw a perpendicular to the x-axis and intersecting the tangent at T. Then draw a line through P, parallel to the x-axis and intersecting the perpendicular through Q at R. Let θ be the inclination of the tangent PT.

P

Q

T

Rθ

From Analytic Geometry, we know that slope of PT = tan θBut triangle PRT, we see that

xRT

PRRT

tan∆

θ ==

However, Δx=dx by DEFINITION 1 . Hence

dxRT

tan =θ

But the derivative of y=f(x) at point P is equal to the slope of the tangent line at that same point P. slope of PT = f’(x)Hence,

( )dxRT

x' f =

And , RT = f’(x) dxBut, dy = f’ (x) dxHence, RT = dy

We see that dy is the increment of the ordinate of the tangent line corresponding to an increment in Δx in x whereas Δy is the corresponding increment of the curve for the same increment in x. We also note that the derivative dy/dx or f’(x) gives the slope of the tangent while the differential dy gives the rise of the tangent line.

DIFFERENTIAL FORMULAS Since we have already considered dy/dx as the ratio of two differentials, then the differentiation formulas may now be expressed in terms of differentials by multiplying both sides of the equation by dx. Thus d(c) = 0 d(x) =dx d(cu) = cdu d(u + v) = du + dv d(uv) = udv + vdu d(u/v) = (vdu – udv)/v2 d(un) = nun-1 du ( ) u2duud =

EXAMPLE 1: Find dy for y = x3 + 5 x −1. ( )

( ) dx 53xdy

dx5dxx3

1x5xddy

2

2

3

+=

+=

−+=

EXAMPLE 2: Find dy for . 1x3

x2y

−=

( ) ( ) ( ) ( )( )

( ) ( ) 22

2

1x3

2dxdy

1x3

x62x6dy

1x3

3x221x3

1x3x2

ddy

−−=∴⇒

−

−−=

−−−=

−=

dx.by itmultiply

and equation the of member right the of

derivative the get simply we practice,In:Note

EXAMPLE 3: Find dy / dx by means of differentials if xy + sin x = ln y .

( )

( )

( ) ( )

( )1xy

xcosyydxdy

xcosyydxdy

1xy

xcosyydxdy

dxdy

xy

dxdy

xcosyydxdy

xy

dx1

dydx xcosydxydy xy

dydx xcosydxydy xy

ydyy1

dx xcosdx ydy x

dyy1

dx xcosdx ydy x

2

2

2

2

−+−=∴

+−=−

−−=−

=++

=++

=++

=++

=++

( ) ( ).tgy ,tfx equations

c parametrifor ,dx

yd and

dxdy

as suchs,derivative

the find to e procedurthe es investigat lesson This

2

2

==

CHAIN RULE FOR PARAMETRIC EQUATIONS

( ) ( )

dtdx

dxdy

dtd

dx

yd and

dtdxdtdy

dxdy

symbols,In

manner. similara in

found are sderivative Higher .dtdx

to dtdy

of ratio the is curve

c parametrithe on dxdy

derivative the that statesRule Chain The

tgy and tfx

equations c parametritheby defined is curve a Suppose

RULECHAIN HET

2

2

==

==

Find the derivatives of the following parametric equations :

t cot2sint-

2cost

dtdxdtdy

dxdy

tcos2dtdy

and tsin2dtdx

:Solution

sint2 y t, 2cos x .1

−===

=−=

==

3t cot3sin3t-

3cos3t

dtdxdtdy

dxdy

t3cos3dtdy

and t3sin3dtdx

:Solution

3t siny 3t, cos x .2

−===

=−=

==

EXAMPLE :

( ) ( )

( ) ( )

( )( ) ( )2t cot

2t sin2tcos

dtdxdtdy

dxdy

2tcosdtdy

and 2t sindtdx

:Solution

2t siny ,2t cos x .3

+−=+−

+==

+=+−=

+=+=

( )( )

t4sin5tsin2t4cos5tcos2

t4sin5tsin24t4cos5tcos24

t4sin20tsin8t4cos20tcos8

dtdxdtdy

dxdy

t4cos20tcos8dtdy

and t4sin20tsin8dtdx

:Solution

5sin4t-t 8sin y 5cos4t, 8cost x .4

−−−=

−−−=

−−−==

−=

−−=

=+=

2

223

dxyd

find ,tty ,1txIf .5 +=−=

2t31t2

dtdxdtdy

dxdy +==

( )( ) ( ) ( )

( )52

2

24

2

2

2

22

2

t91t2

dxyd

t31

t9t61t22t3

dxyd

dxdt

t31t2

dtd

dxyd

+−=∴

•+−=

+=

2

2

dxyd

find , cos 41y , sin 2xIf .6 θ−=θ=

θ=θθ=

θ

θ= tan2cos2sin4

ddxddy

dxdy

( )

θ=∴

θ•θ=

θ•θ=

θθθ

=

3

2

2

2

2

2

2

2

2

2

2

secdx

yd

secsecdx

ydcos21

sec2dx

yddxd

tan2dd

dxyd

( )( ).0,4 at ty and t4tx

:curve parametric the to s line tangent the Find .7235 =−=

( ) ( )125tt2

125tt

2t

t125t2t

dtdxdtdy

dxdy

line. tangent the of slopethe get can we

that so,dxdy

find and derivative the find to have We

22224 −=

−=

−==

( )( )

( )( )( )

( )( )( )

4x81

yx81

4-y is line tangent secondof equation the thus

81

m is 0,4 at line tangent the of slopethe Therefore

81

12252

2dxdy

,2t at ,Now

4x81

yx81

4-y is line tangent of equation the thus

81

m is 0,4 at line tangent the of slopethe Therefore

81

12252

2dxdy

2, t at

defined not is dxdy

0,t at

2t 0,t

04t ,0t

04ttt4t0

becomes curve the of equation c parametrithe 4,0 at ,Now

2

2

23

2335

+−=→−=→

−=

−=−−−

=−=

+=→=→

=

=−

==

=

±===−=

=−→−=