082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

394 Chapter 11 Techniques of differentiation
Using the chain rule,we obtain
dy
dx = dy / dx
dt dt = 2(1+t) =1+t=1+ x 2 2
Example11.8 Giveny = e t +t,x =t 2 +1,find dy using parametric differentiation.
dx
Solution
dy
dt =et +1
dx
dt = 2t
Hence,
dy
dx = dy / dx
dt dt = et +1
2t
In this example, the derivative is expressed in terms oft. This will always be the case
whent has not been eliminated betweenxandy.
Example11.9 Ifx = sint +cost andy =t 2 −t+1find dy (t =0).
dx
Solution
Hence,
dy
dt =2t−1
dy
dx = 2t−1
cost−sint
Whent = 0, dy
dx = −1
1 = −1.
dx
dt =cost−sint
11.3.2 Implicitdifferentiation
Suppose wearetold that
y 3 +x 3 =5sinx+10cosy
Althoughydependsuponx,itisimpossibletowritetheequationintheformy = f (x).
We say y is expressed implicitly in terms of x. The form y = f (x) is an explicit
expression for y in terms of x. However, given an implicit expression for y it is still
possible to find dy dy
. Usually will be expressed in terms of bothxandy. Essentially,
dx dx
the chain ruleisused when differentiating implicit expressions.
Whencalculating dy
dx weneedtodifferentiateafunctionofy,asopposedtoafunction
ofx. For example, we may need tofind d dx (y4 ).
Example11.10 Find
d
(a)
dx (y4 )
(b)
d
dx (y−3 )