Higher Engineering Mathematics, Sixth Edition

328 Higher Engineering Mathematics

(iv)

dy

dx

=y

{

3

x

+

1

xln2x

− 1 −cotx

}

(v)

dy

dx

=

x^3 ln2x

exsinx

{

3

x

+

1

xln2x

− 1 −cotx

}

Now try the following exercise

Exercise 132 Further problems on

differentiating logarithmic functions

In Problems 1 to 6, use logarithmic differentiation

to differentiate the given functions with respect to

the variable.

1. y=

(x− 2 )(x+ 1 )

(x− 1 )(x+ 3 )

⎡

⎢

⎢

⎣

(x− 2 )(x+ 1 )

(x− 1 )(x+ 3 )

{

1

(x− 2 )

+

1

(x+ 1 )

−

1

(x− 1 )

−

1

(x+ 3 )

}

⎤

⎥

⎥

⎦

2. y=

(x+ 1 )( 2 x+ 1 )^3

(x− 3 )^2 (x+ 2 )^4

⎡

⎢

⎢

⎢

⎣

(x+ 1 )( 2 x+ 1 )^3

(x− 3 )^2 (x+ 2 )^4

{

1

(x+ 1 )

+

6

( 2 x+ 1 )

−

2

(x− 3 )

−

4

(x+ 2 )

}

⎤

⎥

⎥

⎥

⎦

3. y=

( 2 x− 1 )

√

(x+ 2 )

(x− 3 )

√

(x+ 1 )^3

⎡

⎢

⎢

⎢

⎣

( 2 x− 1 )

√

(x+ 2 )

(x− 3 )

√

(x+ 1 )^3

{

2

( 2 x− 1 )

+

1

2 (x+ 2 )

−

1

(x− 3 )

−

3

2 (x+ 1 )

}

⎤

⎥

⎥

⎥

⎦

4. y=

e^2 xcos3x

√

(x− 4 )

[

e^2 xcos3x

√

(x− 4 )

{

2 −3tan3x−

1

2 (x− 4 )

}]

5. y= 3 θsinθcosθ
   [
   3 θsinθcosθ

{

1

θ

+cotθ−tanθ

}]

6. y=

2 x^4 tanx

e^2 xln2x

[

2 x^4 tanx

e^2 xln2x

{

4

x

+

1

sinxcosx

− 2 −

1

xln2x

}]

7. Evaluate

dy

dx

whenx=1given

y=

(x+ 1 )^2

√

( 2 x− 1 )

√

(x+ 3 )^3

[

13

16

]

8. Evaluate

dy

dθ

, correct to 3 significant figures,

whenθ=

π

4

giveny=

2eθsinθ

√

θ^5

[− 6 .71]

31.5 Differentiation of[f(x)]x

Whenever an expression to be differentiated con-

tains a term raised to a power which is itself a function

of the variable, then logarithmicdifferentiation must be

used. For example, the differentiation of expressions

such asxx,(x+ 2 )x,x

√

(x− 1 )andx^3 x+^2 can only be

achieved using logarithmic differentiation.

Problem 5. Determine

dy

dx

giveny=xx.

Taking Napierian logarithms of both sides of

y=xxgives:

lny=lnxx=xlnx, by law (iii) of Section 31.2

Differentiating both sides with respect toxgives:

1

y

dy

dx

=(x)

(

1

x

)

+(lnx)( 1 ), using the product rule

i.e.

1

y

dy

dx

= 1 +lnx,

from which,

dy

dx

=y( 1 +lnx)

i.e.

dy

dx

=xx( 1 +lnx)

Problem 6. Evaluate

dy

dx

whenx=−1given

y=(x+ 2 )x.

Taking Napierian logarithms of both sides of

y=(x+ 2 )xgives:

lny=ln(x+ 2 )x=xln(x+ 2 ),by law (iii)

of Section 31.2