Chapter 31

Logarithmic Differentiation

31.1 Introduction to logarithmic differentiation

With certain functions containing more complicated
products and quotients, differentiation is often made
easier if the logarithm of the function is taken before
differentiating. This technique, called ‘logarithmic
differentiation’is achieved with a knowledge of (i) the
laws of logarithms, (ii) the differential coefficients of
logarithmic functions, and (iii) the differentiation of
implicit functions.

31.2 Laws of logarithms

Three laws of logarithmsmay be expressed as:

(i) log(A×B)=logA+logB

(ii) log

(

A

B

)

=logA−logB

(iii) logAn=nlogA

In calculus, Napierian logarithms (i.e. logarithms to a
base of ‘e’) are invariably used. Thus for two func-
tions f(x)andg(x)the laws of logarithms may be
expressed as:

(i) ln[f(x)·g(x)]=lnf(x)+lng(x)

(ii) ln

(

f(x)

g(x)

)

=lnf(x)−lng(x)

(iii) ln[f(x)]n=nlnf(x)

Taking Napierian logarithms of both sides of the equa-

tiony=

f(x)·g(x)

h(x)

gives:

lny=ln

(

f(x)·g(x)

h(x)

)

which may be simplified using the above laws of

logarithms, giving:

lny=lnf(x)+lng(x)−lnh(x)

This latter form of the equation is often easier to

differentiate.

31.3 Differentiation of logarithmic functions

The differential coefficient of the logarithmic function

lnxis given by:

d

dx

(lnx)=

1

x

More generally, it may be shown that:

d

dx

[lnf(x)]=

f′(x)

f(x)

(1)

For example, ify=ln( 3 x^2 + 2 x− 1 )then,

dy

dx

=

6 x+ 2

3 x^2 + 2 x− 1

Similarly, ify=ln(sin3x)then

dy

dx

=

3cos3x

sin3x

=3cot3x.

Now try the following exercise

Exercise 131 Further problems on

differentiating logarithmic functions

Differentiate the following using the laws for

logarithms.

1. ln( 4 x− 10 )

[

4

4 x− 10

]