322 Higher Engineering Mathematics
Thus
d
dx
(
3 y
2 x
)
=
( 2 x)
d
dx
( 3 y)−( 3 y)
d
dx
( 2 x)
( 2 x)^2
=
( 2 x)
(
3
dy
dx
)
−( 3 y)( 2 )
4 x^2
=
6 x
dy
dx
− 6 y
4 x^2
=
3
2 x^2
(
x
dy
dx
−y
)
Problem 5. Differentiatez=x^2 + 3 xcos3ywith
respect toy.
dz
dy
=
d
dy
(x^2 )+
d
dy
( 3 xcos3y)
= 2 x
dx
dy
+
[
( 3 x)(−3sin3y)+(cos 3y)
(
3
dx
dy
)]
= 2 x
dx
dy
− 9 xsin3y+3cos3y
dx
dy
Now try the following exercise
Exercise 129 Further problems on
differentiating implicit functions involving
products and quotients
- Determine
d
dx
( 3 x^2 y^3 ).
[
3 xy^2
(
3 x
dy
dx
+ 2 y
)]
- Find
d
dx
(
2 y
5 x
)
.
[
2
5 x^2
(
x
dy
dx
−y
)]
- Determine
d
du
(
3 u
4 v
)
.
[
3
4 v^2
(
v−u
dv
du
)]
- Givenz= 3
√
ycos3xfind
dz
dx
.
[
3
(
cos3x
2
√
y
)
dy
dx
− 9
√
ysin3x
]
- Determine
dz
dy
givenz= 2 x^3 lny.
[
2 x^2
(
x
y
+3lny
dx
dy
)]
30.4 Further implicit differentiation
An implicitfunctionsuch as 3x^2 +y^2 − 5 x+y=2, may
be differentiated term by term with respect tox.This
gives:
d
dx
( 3 x^2 )+
d
dx
(y^2 )−
d
dx
( 5 x)+
d
dx
(y)=
d
dx
( 2 )
i.e. 6x+ 2 y
dy
dx
− 5 + 1
dy
dx
= 0 ,
using equation (1) and standard derivatives.
An expression for the derivative
dy
dx
in terms ofxand
ymay be obtained by rearranging this latter equation.
Thus:
( 2 y+ 1 )
dy
dx
= 5 − 6 x
from which,
dy
dx
=
5 − 6 x
2 y+ 1
Problem 6. Given 2y^2 − 5 x^4 − 2 − 7 y^3 =0,
determine
dy
dx
Each term in turn is differentiated with respect tox:
Hence
d
dx
( 2 y^2 )−
d
dx
( 5 x^4 )−
d
dx
( 2 )−
d
dx
( 7 y^3 )
=
d
dx
( 0 )
i.e. 4y
dy
dx
− 20 x^3 − 0 − 21 y^2
dy
dx
= 0
Rearranging gives:
( 4 y− 21 y^2 )
dy
dx
= 20 x^3
i.e.
dy
dx
=
20 x^3
(4y− 21 y^2 )
Problem 7. Determine the values of
dy
dx
when
x=4 given thatx^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 )