The introduction is for a student who might not have seen logarithmic differentiation before: (and yes, this technique is extensively used..for example it is used in the “maximum likelihood function” calculation frequently encountered in statistics)

Suppose you are given, say, and you are told to calculate the derivative?

Calculus texts often offer the technique of logarithmic differentiation: write

Now differentiate both sides:

Now multiply both sides by to obtain

\

And this is correct…sort of. Why I say sort of: what happens, at say, ? The derivative certainly exists there but what about that second factor? Yes, the sin(x) gets cancelled out by the first factor, but AS WRITTEN, there is an oh-so-subtle problem with domains.

You can only substitute only after simplifying ..which one might see as a limit process.

But let’s stop and take a closer look at the whole process: we started with and then took the log of both sides. Where is the log defined? And when does ? You got it: this only works when .

So, on the face of it, is justified only when each .

Why can we get away with ignoring all of this, at least in this case?

Well, here is why:

1. If is a differentiable function then

Yes, this is covered in the derivation of material but here goes: write

Now if we get as usual. If then and so in either case:

as required.

THAT is the workaround for calculating where : just calculate . noting that

Yay! We are almost done! But, what about the cases where at least some of the factors are zero at, say ?

Here, we have to bite the bullet and admit that we cannot take the log of the product where any of the factors have a zero, at that point. But this is what we can prove:

Given is a product of differentiable functions and then

This works out to what we want by cancellation of factors.

Here is one way to proceed with the proof:

1. Suppose are differentiable and . Then and

2. Now suppose are differentiable and . Then and

3.Now apply the above to is a product of differentiable functions and

If then by inductive application of 1.

If then let as in 2. Then by 2, we have Now this quantity is zero unless and . But in this case note that

So there it is. Yes, it works ..with appropriate precautions.