328 Higher Engineering Mathematics
(iv)
d y
dx
= y
3
x
+
1
x ln 2x
− 1 − cot x
(v)
dy
dx
=
x 3 ln 2x
e x sin x
3
x
+
1
x ln 2x
− 1 − cot x
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.
1. 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)
⎤
⎥
⎥
⎦
2. y =
(x + 1)(2x + 1) 3
(x − 3) 2 (x + 2) 4
⎡
⎢
⎢
⎢
⎣
(x + 1)(2x + 1) 3
(x − 3) 2 (x + 2) 4
1
(x + 1)
+
6
(2x + 1)
−
2
(x − 3)
−
4
(x + 2)
⎤
⎥
⎥
⎥
⎦
3. y =
(2x − 1)
√ (x + 2)
(x − 3)
(x + 1) 3
⎡
⎢
⎢
⎢
⎣
(2x − 1)
√ (x + 2)
(x − 3)
(x + 1) 3
2
(2x − 1)
+
1
2(x + 2)
−
1
(x − 3)
−
3
2(x + 1)
⎤
⎥
⎥
⎥
⎦
4. y =
e 2x cos 3x
√ (x − 4)
e 2x cos 3x
√ (x − 4)
2 − 3 tan3x −
1
2(x − 4)
5. y = 3θ sin θ cos θ
3θ sin θ cos θ
1
θ
+ cot θ − tan θ
6. y =
2x 4 tan x
e 2x ln 2x
2x 4 tan x
e 2x ln 2x
4
x
+
1
sin x cos x
− 2 −
1
x ln 2x
7. Evaluate
d y
dx
when x = 1 given
y =
(x + 1) 2 √
(2x − 1)
(x + 3) 3
13
16
8. Evaluate
d y
dθ
, correct to 3 significant figures,
when θ =
π
4
given y =
2e θ sin θ
√ θ 5
[−6.71]
31.5 Differentiation of [ f (x)]
x
Whenever an expression to be differentiated contains a term raised to a power which is itself a function
of the variable, then logarithmic differentiation must be
used. For example, the differentiation of expressions
such as x x , (x + 2) x ,
x
√ (x − 1) and x 3x+2 can only be
achieved using logarithmic differentiation.
Problem 5. Determine
d y
dx
given y = x x .
Taking Napierian logarithms of both sides of
y = x x gives:
ln y = ln x x = x ln x, by law (iii) of Section 31.2
Differentiating both sides with respect to x gives:
1
y
d y
dx
= (x)
1
x
+ (ln x)(1), using the product rule
i.e.
1
y
d y
dx
= 1 + ln x,
from which,
d y
dx
= y(1 + ln x)
i.e.
dy
dx
= x x (1 + ln x)
Problem 6. Evaluate
d y
dx
when x = −1 given
y = (x + 2) x .
Taking Napierian logarithms of both sides of
y = (x + 2) x gives:
ln y = ln(x + 2) x = x ln(x + 2), by law (iii)
of Section 31.2
(iv)
d y
dx
= y
3
x
+
1
x ln 2x
− 1 − cot x
(v)
dy
dx
=
x 3 ln 2x
e x sin x
3
x
+
1
x ln 2x
− 1 − cot x
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.
1. 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)
⎤
⎥
⎥
⎦
2. y =
(x + 1)(2x + 1) 3
(x − 3) 2 (x + 2) 4
⎡
⎢
⎢
⎢
⎣
(x + 1)(2x + 1) 3
(x − 3) 2 (x + 2) 4
1
(x + 1)
+
6
(2x + 1)
−
2
(x − 3)
−
4
(x + 2)
⎤
⎥
⎥
⎥
⎦
3. y =
(2x − 1)
√ (x + 2)
(x − 3)
(x + 1) 3
⎡
⎢
⎢
⎢
⎣
(2x − 1)
√ (x + 2)
(x − 3)
(x + 1) 3
2
(2x − 1)
+
1
2(x + 2)
−
1
(x − 3)
−
3
2(x + 1)
⎤
⎥
⎥
⎥
⎦
4. y =
e 2x cos 3x
√ (x − 4)
e 2x cos 3x
√ (x − 4)
2 − 3 tan3x −
1
2(x − 4)
5. y = 3θ sin θ cos θ
3θ sin θ cos θ
1
θ
+ cot θ − tan θ
6. y =
2x 4 tan x
e 2x ln 2x
2x 4 tan x
e 2x ln 2x
4
x
+
1
sin x cos x
− 2 −
1
x ln 2x
7. Evaluate
d y
dx
when x = 1 given
y =
(x + 1) 2 √
(2x − 1)
(x + 3) 3
13
16
8. Evaluate
d y
dθ
, correct to 3 significant figures,
when θ =
π
4
given y =
2e θ sin θ
√ θ 5
[−6.71]
31.5 Differentiation of [ f (x)]
x
Whenever an expression to be differentiated contains a term raised to a power which is itself a function
of the variable, then logarithmic differentiation must be
used. For example, the differentiation of expressions
such as x x , (x + 2) x ,
x
√ (x − 1) and x 3x+2 can only be
achieved using logarithmic differentiation.
Problem 5. Determine
d y
dx
given y = x x .
Taking Napierian logarithms of both sides of
y = x x gives:
ln y = ln x x = x ln x, by law (iii) of Section 31.2
Differentiating both sides with respect to x gives:
1
y
d y
dx
= (x)
1
x
+ (ln x)(1), using the product rule
i.e.
1
y
d y
dx
= 1 + ln x,
from which,
d y
dx
= y(1 + ln x)
i.e.
dy
dx
= x x (1 + ln x)
Problem 6. Evaluate
d y
dx
when x = −1 given
y = (x + 2) x .
Taking Napierian logarithms of both sides of
y = (x + 2) x gives:
ln y = ln(x + 2) x = x ln(x + 2), by law (iii)
of Section 31.2