326 Higher Engineering Mathematics
2. ln(cos 3x)
[−3 tan 3x]
3. ln(3x 3 + x)
9x 2 + 1
3x 3 + x
4. ln(5x 2 + 10x − 7)
10x + 10
5x 2 + 10x − 7
5. ln 8x
1
x
6. ln(x 2 − 1)
2x
x 2 − 1
7. 3 ln 4x
3
x
8. 2 ln(sin x)
[2 cot x]
9. ln(4x
3
− 6x
2
+ 3x)
12x 2 − 12x + 3
4x 3 − 6x 2 + 3x
31.4 Differentiation of further
logarithmic functions
As explained in Chapter 30, by using the function of a
function rule:
d
dx
(ln y) =
1
y
dy
dx
(2)
Differentiation of an expression such as
y =
(1 + x) 2 √
(x − 1)
x
√
(x + 2)
may be achieved by using the
product and quotient rules of differentiation; however the working would be rather complicated. With
logarithmic differentiation the following procedure is
adopted:
(i) Take Napierian logarithms of both sides of the
equation.
Thus ln y = ln
(1 + x) 2 √
(x − 1)
x
√
(x + 2)
= ln
(1 + x)
2
(x − 1)
1
2
x(x + 2)
1
2
(ii) Apply the laws of logarithms.
Thus ln y = ln(1 + x) 2 + ln(x − 1)
1
2
− ln x − ln(x + 2)
1
2 , by laws (i)
and (ii) of Section 31.2
i.e. ln y = 2 ln(1 + x) +
1
2 ln(x − 1)
− ln x −
1
2 ln(x + 2), by law (iii)
of Section 31.2
(iii) Differentiate each term in turn with respect to x
using equations (1) and (2).
Thus
1
y
d y
dx
=
2
(1 + x)
+
1
2
(x − 1)
−
1
x
−
1
2
(x + 2)
(iv) Rearrange the equation to make
d y
dx
the subject.
Thus
d y
dx
= y
2
(1 + x)
+
1
2(x − 1)
−
1
x
−
1
2(x + 2)
(v) Substitute for y in terms of x.
Thus
dy
dx
=
(1 + x)
2
√
(x − 1)
x
√
(x + 2)
2
(1 + x)
+
1
2(x − 1)
−
1
x
−
1
2(x + 2)
Problem 1. Use logarithmic differentiation to
differentiate y =
(x + 1)(x − 2)
3
(x − 3)
Following the above procedure:
(i) Since y =
(x + 1)(x − 2) 3
(x − 3)
then ln y = ln
(x + 1)(x − 2) 3
(x − 3)
(ii) ln y = ln(x + 1) + ln(x − 2) 3 − ln(x − 3),
by laws (i) and (ii) of Section 31.2,
i.e. ln y = ln(x + 1) + 3 ln(x − 2) − ln(x − 3),
by law (iii) of Section 31.2.
(iii) Differentiating with respect to x gives:
1
y
d y
dx
=
1
(x + 1)
+
3
(x − 2)
−
1
(x − 3)
,
by using equations (1) and (2)
(iv) Rearranging gives:
d y
dx
= y
1
(x + 1)
+
3
(x − 2)
−
1
(x − 3)