Higher Engineering Mathematics, Sixth Edition

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

1. Determine

d

dx

( 3 x^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
   √
   ycos3xfind

dz

dx

.

[

3

(

cos3x

2

√

y

)

dy

dx

− 9

√

ysin3x

]

5. 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 )