Calculus:Finding The Derivative Of Logarithmic Functions

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The derivative of logarithmic functions is the differentiation of logathmicfunctions.
Let us look at how we can obtain a formula that applies to this mathematical process.

In mathematics, specifically in calculus and complex analysis, the logarithmic derivative of a function f is defined by the formula given below:

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where f' is the derivative of f. Intuitively, this is the infinitesimal relative change in f; that is, the infinitesimal absolute change in f, namely f', scaled by the current value of f.

When f is a function f(x) of a real variable x, and takes real, strictly positive values, this is equal to the derivative of ln(f), or the natural logarithm of f. This follows directly from the chain rule.

Many properties of the real logarithm also apply to the logarithmic derivative, even when the function does not take values in the positive reals. For example, since the logarithm of a product is the sum of the logarithms of the factors, we have

Thus, it is true for any function that the logarithmic derivative of a product is the sum of the logarithmic derivatives of the factors (when they are defined).

A corollary to this is that the logarithmic derivative of the reciprocal of a function is the negation of the logarithmic derivative of the function:

just as the logarithm of the reciprocal of a positive real number is the negation of the logarithm of the number.

More generally, the logarithmic derivative of a quotient is the difference of the logarithmic derivatives of the dividend and the divisor:

just as the logarithm of a quotient is the difference of the logarithms of the dividend and the divisor.

Generalising in another direction, the logarithmic derivative of a power (with constant real exponent) is the product of the exponent and the logarithmic derivative of the base:

just as the logarithm of a power is the product of the exponent and the logarithm of the base.

In summary, both derivatives and logarithms have a product rule, a reciprocal rule, a quotient rule, and a power rule(compare the list of logarithmic identities); each pair of rules is related through the logarithmic derivative.

Worked Examples:

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Worked Example 5:
Obtain completely the derivative of 

x^2 ln x 

Solution:

According to the product rule, if h(x) is a function of the product of two functions, f(x) and g(x), then:

h(x) = f(x) * g(x)

h ' (x) = [ f ' (x) * g(x) ] + [ g ' (x) * f(x) ]

What this basically says is that the derivative of the product is the derivative of the first function times the second function plus the derivative of second function times the first function.

NOTE: "times" means multiplication

So, let f(x) = x² and g(x) = ln(x)

Therefore,

h(x) = x² * ln(x)


Now, we use the product rule.

h ' (x) = [ f ' (x) * g(x) ] + [ g ' (x) * f(x) ]

First, we do the calculations required in the first pair of brackets:

[ f ' (x) * g(x) ]

From the power rule of derivatives, the derivative of x² is 2x. We leave g(x) = ln(x) alone.

Thus,

[ f ' (x) * g(x) ] = [ 2x * ln(x) ]


Now, we do the calculations required in the second pair of brackets:

[ g ' (x) * f(x) ]

The derivative of ln(x) is 1/x. We leave f(x) = x² alone.

Thus,


[ g ' (x) * f(x) ] = [ (1 / x) * (x²) ]


Now, we put these together to get:


h ' (x) = [ f ' (x) * g(x) ] + [ g ' (x) * f(x) ]

=> h ' (x) = [ 2x * ln(x) ] + [ (1 / x) * (x²) ]

We can rearrange this to get:

h ' (x) = [ 2x * ln(x) ] + [ (1 / x) * (x²) ]

=> h ' (x) = [ (1 / x) * (x²) ] + [ 2x * ln(x) ] <-- All I did here was rearrange the order of the brackets.


We can simplify the expression in the first pair of brackets:

h ' (x) = [ (1 / x) * (x²) ] + [ 2x * ln(x) ]

=> h ' (x) = [ x ] + [ 2x * ln(x) ] <-- x² / x = x

Now, we can factor out an "x" since it's common in both brackets:

h ' (x) = [ x ] + [ 2x * ln(x) ]

=> h ' (x) = x + 2x * ln(x) <-- All I did here was to remove the brackets so that factoring could be seen more clearly.

=> h ' (x) = x [ 1 + 2ln (x) ] .

Worked Example 6:
 Use logarithmic differentiation to find the derivative of the function

y = x ^ (9 cos x) 

 

Solution:

 Taking log both sides, we get, 

 

log y = 9 cosx logx 

now differentiate with respect to x... 

1/y (dy/dx)= 9 {cosx 1/x + logx (-sinx)} 

dy/dx= [ 9 {cosx 1/x + logx (-sinx)}] x ^ (9 cos x).

You can see from the above four examples that good knowledge of indices is required to solve calculus problems.This implies that you must revise the topic of indices as well as surds in order to find calculus easy and interesting.

All the basic laws of indices have been treated in another post on this blog.It will be nice and helpful to give good part of your time to that section.

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