Higher Engineering Mathematics, Sixth Edition

Differentiation of implicit functions 321

du

dx

=

du

dy

×

dy

dx

=

d

dy

(4ln5y)×

dy

dx

=

4

y

dy

dx

(b) Letu=

1

5

e^3 θ−^2 , then, by the functionofa function

rule:

du

dx

=

du

dθ

×

dθ

dx

=

d

dθ

(

1

5

e^3 θ−^2

)

×

dθ

dx

=

3

5

e^3 θ−^2

dθ

dx

Now try the following exercise

Exercise 128 Further problems on

differentiating implicit functions

In Problems 1 and 2 differentiate the given func-

tions with respect tox.

1. (a) 3y^5 (b) 2cos4θ (c)

√ k ⎡ ⎢ ⎢ ⎣

(a) 15 y^4

dy

dx

(b)−8sin4θ

dθ

dx

(c)

1

2

√

k

dk

dx

⎤

⎥

⎥

⎦

2. (a)

5

2

ln3t (b)

3

4

e^2 y+^1 (c)2tan3y

⎡

⎢

⎣

(a)

5

2 t

dt

dx

(b)

3

2

e^2 y+^1

dy

dx

(c)6sec^23 y

dy

dx

⎤

⎥

⎦

3. Differentiate the following with respect toy:
   (a)3sin2θ(b) 4

√

x^3 (c)

2

et

⎡

⎢

⎢

⎣

(a)6cos2θ

dθ

dy

(b) 6

√

x

dx

dy

(c)

− 2

et

dt

dy

⎤

⎥

⎥

⎦

4. Differentiate the following with respect tou:

(a)

2

( 3 x+ 1 )

(b)3sec2θ(c)

2

√

y

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

(a)

− 6

( 3 x+ 1 )^2

dx

du

(b)6sec2θtan2θ

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

dx

(x^2 ),

by the product rule

=(x^2 )

(

1

dy

dx

)

+y( 2 x),

by using equation (1)

=x^2

dy

dx

+ 2 xy

Problem 3. Determine

d

dx

( 2 x^3 y^2 ).

In the product rule of differentiation letu= 2 x^3 and

v=y^2.

Thus

d

dx

( 2 x^3 y^2 )=( 2 x^3 )

d

dx

(y^2 )+(y^2 )

d

dx

( 2 x^3 )

=( 2 x^3 )

(

2 y

dy

dx

)

+(y^2 )( 6 x^2 )

= 4 x^3 y

dy

dx

+ 6 x^2 y^2

= 2 x^2 y

(

2 x

dy

dx

+ 3 y

)

Problem 4. Find

d

dx

(

3 y

2 x

)

.

In the quotient rule of differentiation letu= 3 yand

v= 2 x.