Higher Engineering Mathematics

DIFFERENTIATION OF IMPLICIT FUNCTIONS 321

G

dz

dy

=

d

dy

(x^2 )+

d

dy

(3xcos 3y)

= 2 x

dx

dy

+

[

(3x)(−3 sin 3y)+( cos 3y)

(

3

dx

dy

)]

= 2 x

dx

dy

− 9 xsin 3y+3 cos 3y

dx

dy

Now try the following exercise.

Exercise 131 Further problems on differen-

tiating implicit functions involving products

and quotients

1. Determine

d

dx

(3x^2 y^3 )

[

3 xy^2

(

3 x

dy

dx

+ 2 y

)]

2. Find

d

dx

(

2 y

5 x

)[

2

5 x^2

(

x

dy

dx

−y

)]

3. Determine

d

du

(

3 u

4 v

)[

3

4 v^2

(

v−u

dv

du

)]

4. Givenz= 3

√

ycos 3xfind

dz

[ dx

3

(

cos 3x

2

√

y

)

dy

dx

− 9

√

ysin 3x

]

5. Determine

dz

dy

givenz= 2 x^3 lny

[

2 x^2

(

x

y

+3lny

dx

dy

)]

30.4 Further implicit differentiation

An implicit function such as 3x^2 +y^2 − 5 x+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. 6 x+ 2 y

dy

dx

− 5 + 1

dy

dx

=0,

using equation (1) and standard derivatives.

An expression for the derivative

dy

dx

in terms of

xandymay be obtained by rearranging this latter

equation. Thus:

(2y+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

(2y^2 )−

d

dx

(5x^4 )−

d

dx

(2)−

d

dx

(7y^3 )

=

d

dx

(0)

i.e. 4 y

dy

dx

− 20 x^3 − 0 − 21 y^2

dy

dx

= 0

Rearranging gives:

(4y− 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 tox

gives:

d

dx

(x^2 )+

d

dx

(y^2 )=

d

dx

(25)

i.e. 2 x+ 2 y

dy

dx

= 0

Hence

dy

dx

=−

2 x

2 y

=−

x

y

Sincex^2 +y^2 =25, whenx=4,y=

√

(25− 42 )=± 3

Thus whenx=4 andy=±3,

dy

dx

=−

4

± 3

=±

4

3