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

10.7 Common derivatives 373
Table10.1
Derivatives ofcommonly usedfunctions.
Function,y(x) Derivative,y ′
constant 0
x n
e x
e −x
e ax
lnx
sinx
cosx
sin(ax +b)
cos(ax +b)
tan(ax +b)
cosec(ax +b)
sec(ax +b)
cot(ax +b)
sin −1 (ax +b)
nx n−1
e x
−e −x
ae ax
1
x
cosx
−sinx
acos(ax +b)
−asin(ax +b)
asec 2 (ax +b)
−acosec(ax +b)cot(ax +b)
asec(ax +b)tan(ax +b)
−acosec 2 (ax +b)
a
√
1−(ax+b) 2
Function,y(x) Derivative,y ′
cos −1 (ax +b)
tan −1 (ax +b)
sinh(ax +b)
cosh(ax +b)
tanh(ax +b)
cosech(ax +b)
sech(ax +b)
coth(ax +b)
sinh −1 (ax +b)
cosh −1 (ax +b)
tanh −1 (ax +b)
−a
√
1−(ax+b) 2
a
1+(ax+b) 2
acosh(ax +b)
asinh(ax +b)
asech 2 (ax +b)
−acosech(ax +b)×
coth(ax +b)
−asech(ax +b)×
tanh(ax +b)
−acosech 2 (ax +b)
a
√
(ax+b) 2 +1
a
√
(ax+b) 2 −1
a
1−(ax+b) 2
Solution (a) FromTable 10.1,wefind thatif
y = e ax then y ′ =ae ax
Inthis case,a = −7 and so if
y = e −7x then y ′ = −7e −7x
(b) FromTable 10.1,wefind thatif
y =x n then y ′ =nx n−1
Inthis case,n = 5 and so if
y =x 5 then y ′ = 5x 4
(c) Ify = tan(ax+b)theny ′ =asec 2 (ax+b).Inthiscase,a = 3andb = −2.Henceif
y =tan(3x −2) then y ′ =3sec 2 (3x −2)
(d) Ify = sin(ax +b)theny ′ =acos(ax +b).Herea = ωandb = φ,andsoif
(e) Note that
y = sin(ωx + φ) then y ′ = ωcos(ωx + φ)
1
√ x
= x −1/2 . From Table 10.1 we find that ify = x n theny ′ = nx n−1 . In
this case,n = −1/2 and so if
y = 1 √ x
then y ′ = − 1 2 x−3/2