296 Higher Engineering Mathematics
Using the function of a function rule,
dy
dx
=
dy
du
×
du
dx
=
(
1
2
√
u
)
( 6 x+ 4 )=
3 x+ 2
√
u
i.e.
dy
dx
=
3 x+ 2
√
( 3 x^2 + 4 x− 1 )
Problem 22. Differentiatey=3tan^43 x.
Letu=tan3xtheny= 3 u^4
Hence
du
dx
=3sec^23 x,(from Problem 15), and
dy
du
= 12 u^3
Then
dy
dx
=
dy
du
×
du
dx
=( 12 u^3 )(3sec^23 x)
= 12 (tan3x)^3 (3sec^23 x)
i.e.
dy
dx
=36 tan^33 xsec^23 x
Problem 23. Find the differential coefficient of
y=
2
( 2 t^3 − 5 )^4
y=
2
( 2 t^3 − 5 )^4
= 2 ( 2 t^3 − 5 )−^4 .Letu=( 2 t^3 − 5 ),then
y= 2 u−^4
Hence
du
dt
= 6 t^2 and
dy
du
=− 8 u−^5 =
− 8
u^5
Then
dy
dt
=
dy
du
×
du
dt
=
(
− 8
u^5
)
( 6 t^2 )
=
− 48 t^2
(2t^3 −5)^5
Now try the following exercise
Exercise 118 Further problems on the
function of a function
In Problems 1 to 9, find the differential coefficients
with respect to the variable.
- ( 2 x− 1 )^6 [12( 2 x− 1 )^5 ]
- ( 2 x^3 − 5 x)^5 [5( 6 x^2 − 5 )( 2 x^3 − 5 x)^4 ]
3. 2sin( 3 θ− 2 ) [6cos( 3 θ− 2 )]
4. 2cos^5 α [−10cos^4 αsinα]
5.
1
(x^3 − 2 x+ 1 )^5
[
5 ( 2 − 3 x^2 )
(x^3 − 2 x+ 1 )^6
]
- 5e^2 t+^1 [10e^2 t+^1 ]
- 2cot( 5 t^2 + 3 ) [− 20 tcosec^2 ( 5 t^2 + 3 )]
- 6tan( 3 y+ 1 ) [18sec^2 ( 3 y+ 1 )]
- 2etanθ [2sec^2 θetanθ]
- Differentiateθsin
(
θ−
π
3
)
with respect toθ,
and evaluate, correct to 3 significant figures,
whenθ=
π
2
. [1.86]
27.8 Successive differentiation
When a functiony=f(x)is differentiated with respect
toxthedifferentialcoefficient is writtenas
dy
dx
orf′(x).
If the expression is differentiated again, the second dif-
ferential coefficient is obtained and is written as
d^2 y
dx^2
(pronounced dee twoyby deex squared) or f′′(x)
(pronouncedfdouble-dashx).
By successive differentiation further higher derivatives
such as
d^3 y
dx^3
and
d^4 y
dx^4
may be obtained.
Thus ify= 3 x^4 ,
dy
dx
= 12 x^3 ,
d^2 y
dx^2
= 36 x^2 ,
d^3 y
dx^3
= 72 x,
d^4 y
dx^4
=72 and
d^5 y
dx^5
=0.
Problem 24. Iff(x)= 2 x^5 − 4 x^3 + 3 x−5, find
f′′(x).
f(x)= 2 x^5 − 4 x^3 + 3 x− 5
f′(x)= 10 x^4 − 12 x^2 + 3
f′′(x)= 40 x^3 − 24 x= 4 x(10x^2 −6)