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.
- y=x^2 x [2x^2 x( 1 +lnx)]
- y=( 2 x−[ 1 )x
( 2 x− 1 )x
{
2 x
2 x− 1
+ln( 2 x− 1 )
}]
- y=x
√
(x[+ 3 )
√x(x+ 3 )
{
1
x(x+ 3 )
−
ln(x+ 3 )
x^2
}]
- y= 3 x^4 x+^1
[
3 x^4 x+^1
{
4 +
1
x
+4lnx
}]
- Show that wheny= 2 xxandx= 1 ,
dy
dx
=2.
- Evaluate
d
dx
{√x
(x− 2 )
}
whenx=3.
[
1
3
]
- Show that if y=θθ andθ= 2 ,
dy
dθ
= 6 .77,
correct to 3 significant figures.