Higher Engineering Mathematics

LOGARITHMIC DIFFERENTIATION 327

G

4.y=

e^2 xcos 3x

√

(x−4)

[

e^2 xcos 3x

√

(x−4)

{

2 −3 tan 3x−

1

2(x−4)

}]

5.y= 3 θsinθcosθ

[

3 θsinθcosθ

{

1

θ

+cotθ−tanθ

}]

6.y=

2 x^4 tanx

e^2 xln 2x

[

2 x^4 tanx

e^2 xln 2x

{

4

x

+

1

sinxcosx

− 2 −

1

xln 2x

}]

7. Evaluate

dy

dx

whenx=1given

y=

(x+1)^2

√

(2x−1)

√

(x+3)^3

[

13

16

]

8. Evaluate

dy

dθ

, correct to 3 significant figures,

whenθ=

π

4

giveny=

2eθsinθ

√

θ^5

[− 6 .71]

31.4 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 logarithmic differen-
tiation must be used. For example, the differentia-
tion 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

Differentiating both sides with respect toxgives:

1

y

dy

dx

=(x)

(

1

x+ 2

)

+[ln(x+2)](1),

by the product rule.

Hence

dy

dx

=y

(

x

x+ 2

+ln(x+2)

)

=(x+ 2 )x

{

x

x+ 2

+ln(x+ 2 )

}

Whenx=−1,

dy

dx

=(1)−^1

(

− 1

1

+ln 1

)

=(+1)(−1)=− 1

Problem 7. Determine (a) the differential coef-

ficient of y=x

√

(x−1) and (b) evaluate

dy

dx

whenx=2.

(a)y= x

√

(x−1)=(x−1)

1

x, since by the laws of

indicesn

√

am=a

m

n

Taking Napierian logarithms of both sides gives:

lny=ln(x−1)

1

x=

1

x

ln(x−1),

by law (iii) of Section 31.2.

Differentiating each side with respect toxgives:

1

y

dy

dx

=

(

1

x

)(

1

x− 1

)

+[ln(x−1)]

(

− 1

x^2

)

,

by the product rule.

Hence

dy

dx

=y

{

1

x(x−1)

−

ln(x−1)

x^2

}