Logarithmic Differentiation 327
(v) Substituting for y gives:
dy
dx
=
(x + 1)(x − 2)
3
(x − 3)
1
(x + 1)
+
3
(x − 2)
−
1
(x − 3)
Problem 2. Differentiate y =
(x − 2) 3
(x + 1) 2 (2x − 1)
with respect to x and evaluate
d y
dx
when x = 3.
Using logarithmic differentiation and following the
above procedure:
(i) Since y =
(x − 2) 3
(x + 1) 2 (2x − 1)
then ln y = ln
(x − 2) 3
(x + 1) 2 (2x − 1)
= ln
(x − 2)
3
2
(x + 1) 2 (2x − 1)
(ii) ln y = ln(x − 2)
3
2 − ln(x + 1) 2 − ln(2x − 1)
i.e. ln y =
3
2 ln(x − 2) − 2 ln(x + 1)
− ln(2x − 1)
(iii)
1
y
d y
dx
=
3
2
(x − 2)
−
2
(x + 1)
−
2
(2x − 1)
(iv)
d y
dx
= y
3
2(x − 2)
−
2
(x + 1)
−
2
(2x − 1)
(v)
dy
dx
=
(x − 2) 3
(x + 1)
2 (2x − 1)
3
2(x − 2)
−
2
(x + 1)
−
2
(2x − 1)
When x = 3,
d y
dx
=
(1) 3
(4) 2 (5)
3
2
−
2
4
−
2
5
= ±
1
80
3
5
=±
3
400
or ±0.0075
Problem 3. Given y =
3e 2θ sec 2θ
√
(θ − 2)
determine
d y
dθ
Using logarithmic differentiation and following the
procedure gives:
(i) Since y =
3e 2θ sec 2θ
√
(θ − 2)
then ln y = ln
3e 2θ sec 2θ
√ (θ − 2)
= ln
3e 2θ sec 2θ
(θ − 2)
1
2
(ii) ln y = ln 3e
2θ
+ ln sec 2θ − ln(θ − 2)
1
2
i.e. ln y = ln 3 + ln e
2θ
+ ln sec 2θ
−
1
2 ln(θ − 2)
i.e. ln y = ln 3 + 2θ + ln sec 2θ −
1
2 ln(θ − 2)
(iii) Differentiating with respect to θ gives:
1
y
d y
dθ
= 0 + 2 +
2 sec2θ tan 2θ
sec 2θ
−
1
2
(θ − 2)
from equations (1) and (2)
(iv) Rearranging gives:
d y
dθ
= y
2 + 2 tan 2θ −
1
2(θ − 2)
(v) Substituting for y gives:
dy
dθ
=
3e
2θ sec 2θ
√
(θ − 2)
2 + 2 tan 2θ −
1
2(θ − 2)
Problem 4. Differentiate y =
x 3 ln 2x
e x sin x
with
respect to x.
Using logarithmic differentiation and following the
procedure gives:
(i) ln y = ln
x 3 ln 2x
e x sin x
(ii) ln y = ln x
3
+ ln(ln 2x) − ln(e
x
) − ln(sin x)
i.e. ln y = 3 ln x + ln(ln 2x) − x − ln(sin x)
(iii)
1
y
d y
dx
=
3
x
+
1
x
ln 2x
− 1 −
cos x
sin x
(v) Substituting for y gives:
dy
dx
=
(x + 1)(x − 2)
3
(x − 3)
1
(x + 1)
+
3
(x − 2)
−
1
(x − 3)
Problem 2. Differentiate y =
(x − 2) 3
(x + 1) 2 (2x − 1)
with respect to x and evaluate
d y
dx
when x = 3.
Using logarithmic differentiation and following the
above procedure:
(i) Since y =
(x − 2) 3
(x + 1) 2 (2x − 1)
then ln y = ln
(x − 2) 3
(x + 1) 2 (2x − 1)
= ln
(x − 2)
3
2
(x + 1) 2 (2x − 1)
(ii) ln y = ln(x − 2)
3
2 − ln(x + 1) 2 − ln(2x − 1)
i.e. ln y =
3
2 ln(x − 2) − 2 ln(x + 1)
− ln(2x − 1)
(iii)
1
y
d y
dx
=
3
2
(x − 2)
−
2
(x + 1)
−
2
(2x − 1)
(iv)
d y
dx
= y
3
2(x − 2)
−
2
(x + 1)
−
2
(2x − 1)
(v)
dy
dx
=
(x − 2) 3
(x + 1)
2 (2x − 1)
3
2(x − 2)
−
2
(x + 1)
−
2
(2x − 1)
When x = 3,
d y
dx
=
(1) 3
(4) 2 (5)
3
2
−
2
4
−
2
5
= ±
1
80
3
5
=±
3
400
or ±0.0075
Problem 3. Given y =
3e 2θ sec 2θ
√
(θ − 2)
determine
d y
dθ
Using logarithmic differentiation and following the
procedure gives:
(i) Since y =
3e 2θ sec 2θ
√
(θ − 2)
then ln y = ln
3e 2θ sec 2θ
√ (θ − 2)
= ln
3e 2θ sec 2θ
(θ − 2)
1
2
(ii) ln y = ln 3e
2θ
+ ln sec 2θ − ln(θ − 2)
1
2
i.e. ln y = ln 3 + ln e
2θ
+ ln sec 2θ
−
1
2 ln(θ − 2)
i.e. ln y = ln 3 + 2θ + ln sec 2θ −
1
2 ln(θ − 2)
(iii) Differentiating with respect to θ gives:
1
y
d y
dθ
= 0 + 2 +
2 sec2θ tan 2θ
sec 2θ
−
1
2
(θ − 2)
from equations (1) and (2)
(iv) Rearranging gives:
d y
dθ
= y
2 + 2 tan 2θ −
1
2(θ − 2)
(v) Substituting for y gives:
dy
dθ
=
3e
2θ sec 2θ
√
(θ − 2)
2 + 2 tan 2θ −
1
2(θ − 2)
Problem 4. Differentiate y =
x 3 ln 2x
e x sin x
with
respect to x.
Using logarithmic differentiation and following the
procedure gives:
(i) ln y = ln
x 3 ln 2x
e x sin x
(ii) ln y = ln x
3
+ ln(ln 2x) − ln(e
x
) − ln(sin x)
i.e. ln y = 3 ln x + ln(ln 2x) − x − ln(sin x)
(iii)
1
y
d y
dx
=
3
x
+
1
x
ln 2x
− 1 −
cos x
sin x