322 Higher Engineering Mathematics
Thus
d
dx
3y
2x
=
(2x)
d
dx
(3y) − (3y)
d
dx
(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.
dz
dy
=
d
dy
(x
2
) +
d
dy
(3x cos 3y)
= 2x
dx
dy
+
(3x)(−3 sin3y) + (cos 3y)
3
dx
dy
= 2x
dx
dy
− 9x sin 3y +3 cos 3y
dx
dy
Now try the following exercise
Exercise 129 Further problems on
differentiating implicit functions involving
products and quotients
1. Determine
d
dx
(3x
2 y
3
).
3x y 2
3x
dy
dx
+ 2y
2. Find
d
dx
2y
5x
.
2
5x 2
x
dy
dx
− y
3. Determine
d
du
3u
4v
.
3
4v 2
v − u
dv
du
4. Given z = 3
√ y cos 3x find
dz
dx
.
3
cos 3x
2
√ y
dy
dx
− 9
√ y sin 3x
5. Determine
dz
dy
given z = 2x
3 ln y.
2x
2
x
y
+ 3 ln y
dx
dy
30.4 Further implicit differentiation
An implicit function such as 3x 2 + y 2 − 5x + y = 2, may
be differentiated term by term with respect to x. This
gives:
d
dx
(3x
2
) +
d
dx
(y
2
) −
d
dx
(5x) +
d
dx
(y) =
d
dx
(2)
i.e.
6x + 2y
dy
dx
− 5 + 1
dy
dx
= 0,
using equation (1) and standard derivatives.
An expression for the derivative
dy
dx
in terms of x and
y may be obtained by rearranging this latter equation.
Thus:
(2y + 1)
dy
dx
= 5 − 6x
from which,
dy
dx
=
5 −6x
2y + 1
Problem 6. Given 2y 2 − 5x 4 − 2 − 7y 3 = 0,
determine
dy
dx
Each term in turn is differentiated with respect to x:
Hence
d
dx
(2y
2
) −
d
dx
(5x
4
) −
d
dx
(2) −
d
dx
(7y
3
)
=
d
dx
(0)
i.e.
4y
dy
dx
− 20x
3
− 0 − 21y
2 dy
dx
= 0
Rearranging gives:
(4y − 21y
2
)
dy
dx
= 20x
3
i.e.
dy
dx
=
20x 3
(4y − 21y 2 )
Problem 7. Determine the values of
dy
dx
when
x = 4 given that x 2 + y 2 = 25.
Differentiating each term in turn with respect to x
gives:
d
dx
(x
2
) +
d
dx
(y
2
) =
d
dx
(25)