differentiation

320 DIFFERENTIAL CALCULUS
Now try the following exercise.
Exercise 130 Further problems on differentiating
implicit functions
In Problems 1 and 2 differentiate the given functions
with respect to x.
1. (a) 3y 5 (b) 2 cos 4θ (c) √ k
⎡
(a) 15y 4 dy (b) −8 sin 4θ dθ ⎤
⎢ dx
dx
⎥
⎣ 1
(c)
2 √ dk
⎦
k dx
2. (a) 5 2 ln 3t (b) 3 4 e2y+1 (c) 2 tan 3y
⎡
(a) 5 dt
(b) 3 dy
⎤
e2y+1
⎢ 2t dx
2 dx ⎥
⎣
(c) 6 sec 2 3y dy
⎦
dx
3. Differentiate the following with respect to y:
(a) 3 sin 2θ (b) 4 √ x 3 (c) 2
⎡
e t
(a) 6 cos 2θ dθ
⎢
dy
⎣
(c) −2 dt
e t dy
(b) 6 √ x dx
dy
4. Differentiate the following with respect to u:
(a)
2
(3x + 1)
2
(b) 3 sec 2θ (c) √ y
⎡
−6 dx
(a)
(3x + 1) 2 du
(b) 6 sec 2θ tan 2θ dθ
⎢
du
⎣
(c) √ −1 dy
y 3 du
30.3 Differentiating implicit
functions containing products
and quotients
⎤
⎥
⎦
⎤
⎥
⎦
The product and quotient rules of differentiation
must be applied when differentiating functions containing
products and quotients of two variables.
For example,
d
dx (x2 y) = (x 2 ) d dx (y) + (y) d dx (x2 ),
by the product rule
(
= (x 2 ) 1 dy
dx
)
+ y(2x),
by using equation (1)
= x 2 dy
dx + 2xy
Problem 3. Determine d dx (2x3 y 2 ).
In the product rule of differentiation let u = 2x 3 and
v = y 2 .
Thus d dx (2x3 y 2 ) = (2x 3 ) d dx (y2 ) + (y 2 ) d dx (2x3 )
(
= (2x 3 ) 2y dy )
+ (y 2 )(6x 2 )
dx
= 4x 3 y dy
dx + 6x2 y 2
(
= 2x 2 y 2x dy )
dx + 3y
Problem 4.
Find d dx
( ) 3y
.
2x
In the quotient rule of differentiation let u = 3y and
v = 2x.
Thus d dx
( ) 3y
(2x) d
=
dx (3y) − (3y) d dx (2x)
2x
(2x) 2
=
=
(
(2x) 3 dy
dx
)
− (3y)(2)
4x 2
6x dy
dx − 6y
4x 2 = 3
2x 2 (
x dy
dx − y )
Problem 5. Differentiate z = x 2 + 3x cos 3y
with respect to y.