Higher Engineering Mathematics

Differential calculus

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 coef-

ficients of logarithmic functions, and (iii) the differ-

entiation 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-

tionsf(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

equationy=

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 func-

tion 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(3x^2 + 2 x−1) then,

dy

dx

=

6 x+ 2

3 x^2 + 2 x− 1

Similarly, ify=ln(sin 3x) then

dy

dx

=

3 cos 3x

sin 3x

=3 cot 3x.

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 as

y=

(1+x)^2

√

(x−1)

x

√

(x+2)

may be achieved by using the

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

}