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

11.3 Parametric, implicit and logarithmic differentiation 397
(b) Consider d dx (x2 y 3 ). Using the product rule wefind
d
dx (x2 y 3 ) = d dx (x2 )y 3 +x 2 d dx (y3 ) = 2xy 3 +x 2 3y 2dy
dx
Consider d dx (ey ).Letz = e y so dz
dy = ey . Hence,
d
dx (ey )= dz
dx = dz
dy
dy
dx = eydy dx
So, upon differentiating, the equation becomes
So,
2xy 3 +x 2 3y 2dy
dx −eydy dx = 2e2x
dy
dx (3x2 y 2 −e y ) = 2e 2x −2xy 3
from which
dy
dx = 2e2x −2xy 3
3x 2 y 2 −e y
11.3.3 Logarithmicdifferentiation
Thetechniqueof logarithmicdifferentiationisusefulwhenweneedtodifferentiatea
cumbersomeproduct.The methodinvolves takingthe naturallogarithm ofthe function
tobedifferentiated.This isillustratedinthe following examples.
Example11.15 Given thaty =t 2 (1−t) 8 find dy
dt .
Solution Theproductrulecouldbeusedbutwewilldemonstrateanalternativetechnique.Taking
the natural logarithm of both sides ofthe given equation yields
lny = ln(t 2 (1−t) 8 )
Usingthe laws of logarithms we can write thisas
lny=lnt 2 +ln(1−t) 8
=2lnt+8ln(1−t)
Both sides of thisequation are now differentiated w.r.t.t togive
d
dt (lny) = d dt (2lnt)+ d (8ln(1 −t))
dt
The evaluation of d (lny) has already been found inExample 11.11, and so
dt
1dy
y dt = 2 t − 8
1 −t