## Differentiate Logs with Proof

In order to differentiate logs we must use the chain rule. The simplest type of log to differentiate is a natural log this can be done as shown below.

**Differentiate Natural Logs**

A natural log is a log to the base e.

d/dx (ln x) = 1/x

However if we want to differentiate ln(f(x)) we must use the chain rule to get

d/dx (ln(f(x)) = f'(x)/f(x)

**Proof of Derivative of Natural Logs**

Consider

y=ln(x)

then from the definition of a log we get

e^{y} = x –(1)

Differentiate each side with respect to x (you need to use implicit differentiation for the left to get e^{y} dy/dx) to get

e^{y} dy/dx = 1

but from (1) we know that e^{y} = x which we can substitute to get

x dy/dx =1

giving the derivative

dy/dx = 1/x

## Log Laws: Adding and Subtracting Logs

It is often useful to combine several different logs, being added or subtracted, into a single log that can then be manipulated more easily.

For example it is easier to find log(6) than log(3)+log(4) – log(2)

In order to combine logs like this you need to use the following rules:

log_{c}(a) + log_{c}(b) = log(ab)

and

log_{c}(a)-log_{c}(b) = log(a/b)

**Remember:** all logs must be in the same base, you cant use this to add log_{2}(a) + log_{3}b

This rule can be proved quite simply as follows:

Let all logs be to the base c.

if log a = m and log b = n

then a=c^{m} and b = c^{n}

so ab = c^{m}c^{n}= c^{m+n}

by taking logs we obtain

log(ab)=log(c^{m+n}) = m + n

and by substituting in the values of m and n we get the required result

log(ab) = log a + log b

The result for log (a/b) can also be found in this way but I’ll leave that one for you to work out.

By David Woodford

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