Date post: | 22-Jan-2018 |
Category: |
Education |
Upload: | james-tagara |
View: | 151 times |
Download: | 0 times |
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
+−=→−=→
−=
−=−−−
=−=
+=→=→
=
=−
==
=
±===−=
=−→−=