Logarithmic Differentiation 329

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

+ln1

)

=(+ 1 )(− 1 )=− 1

Problem 7. Determine (a) the differential

coefficient ofy=x

√

(x− 1 )and (b) evaluate

dy

dx

whenx=2.

(a) y=x

√

(x− 1 )=(x− 1 )

(^1) x
, since by the laws of
indices n
√
am=a
mn
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
}
i.e.
dy
dx
=x
√
(x− 1 )
{
1
x(x− 1 )
−
ln(x− 1 )
x^2
}
(b) Whenx= 2 ,
dy
dx
=^2
√
( 1 )
{
1
2 ( 1 )
−
ln( 1 )
4
}
=± 1
{
1
2
− 0
}
=±
1
2
Problem 8. Differentiatex^3 x+^2 with respect tox.
Lety=x^3 x+^2
Taking Napierian logarithms of both sides gives:
lny=lnx^3 x+^2
i.e. lny=( 3 x+ 2 )lnx, by law (iii) of Section 31.2.
Differentiating each term with respect toxgives:
1
y
dy
dx
=( 3 x+ 2 )
(
1
x
)
+(lnx)( 3 ),
by the product rule.
Hence
dy
dx
=y
{
3 x+ 2
x
+3lnx
}
=x^3 x+^2
{
3 x+ 2
x
+3lnx
}
=x^3 x+^2
{
3 +
2
x
+3lnx
}
Now try the following exercise
Exercise 133 Further problems on
differentiating[f(x)]xtype functions
In Problems 1 to 4, differentiate with respect tox.

1. y=x^2 x [2x^2 x( 1 +lnx)]


2. y=( 2 x−[ 1 )x
   ( 2 x− 1 )x

{

2 x

2 x− 1

+ln( 2 x− 1 )

}]

3. y=x

√

(x[+ 3 )

√x(x+ 3 )

{

1

x(x+ 3 )

−

ln(x+ 3 )

x^2

}]

4. y= 3 x^4 x+^1

[

3 x^4 x+^1

{

4 +

1

x

+4lnx

}]

5. Show that wheny= 2 xxandx= 1 ,
   dy
   dx

=2.

6. Evaluate

d

dx

{√x

(x− 2 )

}

whenx=3.

[

1

3

]

7. Show that if y=θθ andθ= 2 ,

dy

dθ

= 6 .77,

correct to 3 significant figures.