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.
- 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 )
}
⎤
⎥
⎥
⎦
- 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 )
}
⎤
⎥
⎥
⎥
⎦
- 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 )
}
⎤
⎥
⎥
⎥
⎦
- y=
e^2 xcos3x
√
(x− 4 )
[
e^2 xcos3x
√
(x− 4 )
{
2 −3tan3x−
1
2 (x− 4 )
}]
- y= 3 θsinθcosθ
[
3 θsinθcosθ
{
1
θ
+cotθ−tanθ
}]
- y=
2 x^4 tanx
e^2 xln2x
[
2 x^4 tanx
e^2 xln2x
{
4
x
+
1
sinxcosx
− 2 −
1
xln2x
}]
- Evaluate
dy
dx
whenx=1given
y=
(x+ 1 )^2
√
( 2 x− 1 )
√
(x+ 3 )^3
[
13
16
]
- 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