Logarithmic Differentiation 327

(v) Substituting forygives:

dy

dx

=

(x+ 1 )(x− 2 )^3

(x− 3 )

{

1

(x+ 1 )

+

3

(x− 2 )

−

1

(x− 3 )

}

Problem 2. Differentiatey=

√

(x− 2 )^3

(x+ 1 )^2 ( 2 x− 1 )

with respect toxand evaluate

dy

dx

whenx=3.

Using logarithmic differentiation and following the
above procedure:

(i) Since y=

√

(x− 2 )^3

(x+ 1 )^2 ( 2 x− 1 )

then lny=ln

{ √

(x− 2 )^3

(x+ 1 )^2 ( 2 x− 1 )

}

=ln

{

(x− 2 )

3

2

(x+ 1 )^2 ( 2 x− 1 )

}

(ii) lny=ln(x− 2 )

3

(^2) −ln(x+ 1 )^2 −ln( 2 x− 1 )
i.e. lny=^32 ln(x− 2 )−2ln(x+ 1 )
−ln( 2 x− 1 )
(iii)
1
y
dy
dx

3
2
(x− 2 )
−
2
(x+ 1 )
−
2
( 2 x− 1 )
(iv)
dy
dx
=y
{
3
2 (x− 2 )
−
2
(x+ 1 )
−
2
( 2 x− 1 )
}
(v)
dy
dx

√
(x−2)^3
(x+1)^2 (2x−1)
{
3
2(x−2)
−
2
(x+ 1 )
−
2
( 2 x− 1 )
}
Whenx= 3 ,
dy
dx

√
( 1 )^3
( 4 )^2 ( 5 )
(
3
2
−
2
4
−
2
5
)
=±
1
80
(
3
5
)
=±
3
400
or± 0. 0075
Problem 3. Giveny=
3e^2 θsec2θ
√
(θ− 2 )
determine
dy
dθ
Using logarithmic differentiation and following the
procedure gives:
(i) Since y=
3e^2 θsec2θ
√
(θ− 2 )
then lny=ln
{
3e^2 θsec2θ
√
(θ− 2 )
}
=ln
{
3 e^2 θsec2θ
(θ− 2 )
1
2
}
(ii) lny=ln3e^2 θ+lnsec2θ−ln(θ− 2 )
1
2
i.e. lny=ln3+lne^2 θ+lnsec2θ
−^12 ln(θ− 2 )
i.e. lny=ln3+ 2 θ+lnsec2θ−^12 ln(θ− 2 )
(iii) Differentiating with respect toθgives:
1
y
dy
dθ
= 0 + 2 +
2sec2θtan2θ
sec2θ
−
1
2
(θ− 2 )
from equations (1) and (2)
(iv) Rearranging gives:
dy
dθ
=y
{
2 +2tan2θ−
1
2 (θ− 2 )
}
(v) Substituting forygives:
dy
dθ

3e^2 θsec2θ
√
(θ− 2 )
{
2 +2tan2θ−
1
2 (θ− 2 )
}
Problem 4. Differentiatey=
x^3 ln2x
exsinx
with
respect tox.
Using logarithmic differentiation and following the
procedure gives:
(i) lny=ln
{
x^3 ln2x
exsinx
}
(ii) lny=lnx^3 +ln(ln2x)−ln(ex)−ln(sinx)
i.e. lny=3lnx+ln(ln2x)−x−ln(sinx)
(iii)
1
y
dy
dx

3
x

• 1
  x
  ln2x
  − 1 −
  cosx
  sinx