Higher Engineering Mathematics

320 DIFFERENTIAL CALCULUS

Now try the following exercise.

Exercise 130 Further problems on differen-

tiating implicit functions

In Problems 1 and 2 differentiate the given func-

tions with respect tox.

1. (a) 3y^5 (b) 2 cos 4θ (c)

√ k ⎡ ⎢ ⎢ ⎣

(a) 15y^4

dy

dx

(b)−8 sin 4θ

dθ

dx

(c)

1

2

√

k

dk

dx

⎤

⎥

⎥

⎦

2. (a)

5

2

ln 3t (b)

3

4

e^2 y+^1 (c) 2 tan 3y

⎡

⎢

⎣

(a)

5

2 t

dt

dx

(b)

3

2

e^2 y+^1

dy

dx

(c) 6 sec^23 y

dy

dx

⎤

⎥

⎦

3. Differentiate the following with respect toy:

(a) 3 sin 2θ (b) 4

√

x^3 (c)

2

⎡ et

⎢

⎢

⎣

(a) 6 cos 2θ

dθ

dy

(b) 6

√

x

dx

dy

(c)

− 2

et

dt

dy

⎤

⎥

⎥

⎦

4. Differentiate the following with respect tou:

(a)

2

(3x+1)

(b) 3 sec 2θ (c)

2 √ y ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

(a)

− 6

(3x+1)^2

dx

du

(b) 6 sec 2θtan 2θ

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 con-
taining 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(2x),

by using equation (1)

=x^2

dy

dx

+ 2 xy

Problem 3. Determine

d

dx

(2x^3 y^2 ).

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

v=y^2.

Thus

d

dx

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

d

dx

(y^2 )+(y^2 )

d

dx

(2x^3 )

=(2x^3 )

(

2 y

dy

dx

)

+(y^2 )(6x^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.

Thus

d

dx

(

3 y

2 x

)

=

(2x)

d

dx

(3y)−(3y)

d

dx

(2x)

(2x)^2

=

(2x)

(

3

dy

dx

)

−(3y)(2)

4 x^2

=

6 x

dy

dx

− 6 y

4 x^2

=

3

2 x^2

(

x

dy

dx

−y

)

Problem 5. Differentiate z=x^2 + 3 xcos 3y

with respect toy.