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

398 Chapter 11 Techniques of differentiation
Hence
(
dy 2
dt =y t − 8 )
1 −t
Finally, replacingybyt 2 (1 −t) 8 wehave
(
dy 2
dt =t2 (1−t) 8 t − 8 )
1 −t
Example11.16 Given
find dy
dx .
y =x 3 (1+x) 9 e 6x
Solution Theproductrulecouldbeused.However,wewilluselogarithmicdifferentiation.Taking
the natural logarithm of the equation and applying the laws of logarithms produces
lny = ln(x 3 (1 +x) 9 e 6x ) =lnx 3 +ln(1 +x) 9 +lne 6x
lny=3lnx+9ln(1+x)+6x
This equation isnow differentiated:
and so
1dy
y dx = 3 x + 9
1 +x +6
(
dy 3
dx =y x + 9 )
1 +x +6
= 3x 2 (1 +x) 9 e 6x +9x 3 (1 +x) 8 e 6x +6x 3 (1 +x) 9 e 6x
Example11.17 Ify = √ 1 +t 2 sin 2 t findy ′ .
Solution Taking logarithms we find
lny=ln( √ 1 +t 2 sin 2 t)
= ln √ 1 +t 2 +ln(sin 2 t)
= 1 2 ln(1 +t2 )+2ln(sint)
Differentiation yields
1
y y′ = 1 2t
21 +t +2cost 2 sint
( ) t
y ′ =y
1 +t +2cott 2
= √ 1 +t 2 sin 2 t
( t
1 +t 2 +2cott )