Differentiation of implicit functions 321
du
dx
=
du
dy
×
dy
dx
=
d
dy
(4 ln 5y) ×
dy
dx
=
4
y
dy
dx
(b) Let u =
1
5
e 3θ−2 , then, by the function of a function
rule:
du
dx
=
du
dθ
×
dθ
dx
=
d
dθ
1
5
e
3θ−2
×
dθ
dx
=
3
5
e
3θ −2 dθ
dx
Now try the following exercise
Exercise 128 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
dx
(b) −8 sin4θ
dθ
dx
(c)
1
2
√
k
dk
dx
⎤
⎥
⎥
⎦
2. (a)
5
2
ln 3t (b)
3
4
e 2y+1 (c) 2 tan 3y
⎡
⎢
⎣
(a)
5
2t
dt
dx
(b)
3
2
e 2y+1 dy
dx
(c) 6 sec 2 3y
dy
dx
⎤
⎥
⎦
3. Differentiate the following with respect to y:
(a) 3 sin2θ (b) 4
√
x 3 (c)
2
e t
⎡
⎢
⎢
⎣
(a) 6 cos 2θ
dθ
dy
(b) 6
√
x
dx
dy
(c)
−2
e t
dt
dy
⎤
⎥
⎥
⎦
4. Differentiate the following with respect to u:
(a)
2
(3x + 1)
(b) 3 sec2θ (c)
2
√ y
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
(a)
−6
(3x + 1) 2
dx
du
(b) 6 sec2θ tan 2θ
dθ
du
(c)
−1
y 3
dy
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
(x
2 y) = (x
2
)
d
dx
(y) + (y)
d
d x
(x
2
),
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
(2x 3 y 2 ).
In the product rule of differentiation let u = 2x 3 and
v = y 2 .
Thus
d
dx
(2x
3 y
2
) = (2x
3
)
d
dx
(y
2
) + (y
2
)
d
dx
(2x
3
)
= (2x
3
)
2y
dy
dx
+ (y
2
)(6x
2
)
= 4x
3 y
dy
dx
+ 6x
2 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.