differentiation

296 DIFFERENTIAL CALCULUS
2
y =
(2t 3 − 5) 4 = 2(2t3 − 5) −4 . Let u = (2t 3 − 5),
then y = 2u −4
Hence
Then
du
dt = 6t2 and dy
du =−8u−5 = −8
u 5
dy
dt = dy
du × du ( ) −8
dt = u 5 (6t 2 )
= −48t2
(2t 3 − 5) 5
Now try the following exercise.
Exercise 120 Further problems on the function
of a function
In Problems 1 to 8, find the differential coefficients
with respect to the variable.
1. (2x 3 − 5x) 5 [5(6x 2 − 5)(2x 3 − 5x) 4 ]
2. 2 sin (3θ − 2) [6 cos (3θ − 2)]
3. 2 cos 5 α [−10 cos 4 α sin α]
[
1
5(2 − 3x 2 ]
)
4.
(x 3 − 2x + 1) 5 (x 3 − 2x + 1) 6
5. 5e 2t+1 [10e 2t+1 ]
6. 2 cot (5t 2 + 3) [−20t cosec 2 (5t 2 + 3)]
7. 6 tan (3y + 1) [18 sec 2 (3y + 1)]
8. 2e tan θ [2 sec 2 θ e tan θ ]
(
9. Differentiate θ sin θ − π )
with respect to θ,
and evaluate, correct to 3 significant figures,
when θ = π [1.86]
2
By successive differentiation further higher derivatives
such as d3 y
dx 3 and d4 y
may be obtained.
dx4 Thus if y = 3x 4 , dy
dx = 12x3 , d2 y
dx 2 = 36x2 ,
d 3 y
dx 3 = 72x, d4 y
dx 4 = 72 and d5 y
dx 5 = 0.
Problem 24.
f ′′ (x).
If f (x) = 2x 5 − 4x 3 + 3x − 5, find
f (x) = 2x 5 − 4x 3 + 3x − 5
f ′ (x) = 10x 4 − 12x 2 + 3
f ′′ (x) = 40x 3 − 24x = 4x(10x 2 − 6)
Problem 25. If y = cos x − sin x, evaluate x,in
the range 0 ≤ x ≤ π 2 , when d2 y
is zero.
dx2 Since y = cos x − sin x,
d 2 y
=−cos x + sin x.
dx2 dy
=−sin x − cos x and
dx
When d2 y
is zero, −cos x + sin x = 0,
dx2 i.e. sin x = cos x or sin x
cos x = 1.
Hence tan x = 1 and x = arctan1 = 45 ◦ or π 4 rads
in the range 0 ≤ x ≤ π 2
27.7 Successive differentiation
When a function y = f (x) is differentiated with
respect to x the differential coefficient is written as
dy
dx or f ′ (x). If the expression is differentiated again,
the second differential coefficient is obtained and
is written as d2 y
(pronounced dee two y by dee x
dx2 squared) or f ′′ (x) (pronounced f double-dash x).
Problem 26.
Given y = 2xe −3x show that
d 2 y
dx 2 + 6dy + 9y = 0.
dx
y = 2xe −3x (i.e. a product)
dy
Hence
dx = (2x)(−3e−3x ) + (e −3x )(2)
=−6xe −3x + 2e −3x