326 Higher Engineering Mathematics
- ln(cos 3x) [−3tan3x]
- ln( 3 x^3 +x)
[
9 x^2 + 1
3 x^3 +x]- ln( 5 x^2 + 10 x− 7 )
[
10 x+ 10
5 x^2 + 10 x− 7]- ln8x
[
1
x]- ln(x^2 − 1 )
[
2 x
x^2 − 1]- 3 ln4x
[
3
x]- 2 ln(sinx) [2 cotx]
- ln( 4 x^3 − 6 x^2 + 3 x)
[
12 x^2 − 12 x+ 3
4 x^3 − 6 x^2 + 3 x]31.4 Differentiation of further
logarithmic functions
As explained in Chapter 30, by using the function of a
function rule:
d
dx(lny)=(
1
y)
dy
dx(2)Differentiation of an expression such asy=( 1 +x)^2√
(x− 1 )
x√
(x+ 2 )maybeachievedbyusingthe
product and quotient rules of differentiation; how-
ever 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 lny=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 lny=ln( 1 +x)^2 +ln(x− 1 )1
2−lnx−ln(x+ 2 )(^12)
,bylaws(i)
and (ii) of Section 31.2
i.e. lny=2ln( 1 +x)+^12 ln(x− 1 )
−lnx−^12 ln(x+ 2 ), by law (iii)
of Section 31.2
(iii) Differentiateeach term in turn with respect tox
using equations (1) and (2).
Thus
1
y
dy
dx
2
( 1 +x)
- 1
2
(x− 1 )
−
1
x
−
1
2
(x+ 2 )
(iv) Rearrange the equation to make
dy
dx
the subject.
Thus
dy
dx
=y
{
2
( 1 +x)
1
2 (x− 1 )
−
1
x
−
1
2 (x+ 2 )
}
(v) Substitute foryin terms ofx.
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
differentiatey=
(x+ 1 )(x− 2 )^3
(x− 3 )
Following the above procedure:
(i) Since y=
(x+ 1 )(x− 2 )^3
(x− 3 )
then lny=ln
{
(x+ 1 )(x− 2 )^3
(x− 3 )
}
(ii) lny=ln(x+ 1 )+ln(x− 2 )^3 −ln(x− 3 ),
by laws (i) and (ii) of Section 31.2,
i.e. lny=ln(x+ 1 )+3ln(x− 2 )−ln(x− 3 ),
by law (iii) of Section 31.2.
(iii) Differentiating with respect toxgives:
1
y
dy
dx
1
(x+ 1 )
3
(x− 2 )
−
1
(x− 3 )
,
by using equations (1) and (2)
(iv) Rearranging gives:
dy
dx
=y
{
1
(x+ 1 )
3
(x− 2 )
−
1
(x− 3 )
}