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

(b) Ify = ln(1 −t),theny ′ = −1
1 −t = − 1
1 −t = 1
t −1 .
(c) Ify =8ln(2−3t),theny ′ =8 −3
2−3t = − 24
2−3t = 24
3t−2 .
(d) Ify =ln(1 +x)theny ′ = 1
1 +x .
(e) Ify = ln(1 +cosx)theny ′ = −sinx
1+cosx = − sinx
1+cosx .
11.2 Rules of differentiation 391
Example11.6 Differentiate
(a) y = 3e sinx
(b) y = (3t 2 +2t −9) 10
(c)y= √ 1 +t 2
Solution Inthese examples we mustformulate the functionzourselves.
(a) Letz(x) = sinx. Theny(z) = 3e z so dy
dz = 3ez ;z(x) = sinx and so dz
dx = cosx.
The chain ruleisused tofind dy
dx .
dy
dx = dy
dz × dz
dx = 3ez cosx = 3e sinx cosx
(b) Letz(t) = 3t 2 +2t −9.Theny(z) =z 10 , dy
dz = 10z9 , dz = 6t +2.Usingthechain
dt
rule dy
dt isfound:
dy
dt = dy
dz × dz
dt = 10z9 (6t +2) = 20(3t +1)(3t 2 +2t−9) 9
(c) Letz(t) =1+t 2 .Theny= √ z =z 1/2 , dy
dz = 1 2 z−1/2 and dz
dt = 2t.Usingthechain
rule,we obtain
dy
dt = dy
dz × dz
dt = 1 2 z−1/2 2t = √ t t
= √ z 1 +t
2
EXERCISES11.2
1 Usethe product ruleto differentiate the following
functions:
(a) y=sinxcosx
(b) y=lnttant
(c) y = (t 3 +1)e 2t
(d) y= √ xe x
(e) y=e t sintcost
(f) y = 3sinh2tcosh3t
(g) y=(1+sint)tant
(h) y = 4sinh(t +1)cosh(1 −t)