i.e., is the set of all points such that is defined. The rule for is

and

Thus

So in this case . Thus composition is not a commutative operation.

If and , then

and

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and . Suppose and , and is differentiable at and is differentiable at . Then is differentiable at , and

Hence

Let
. Then where

Hence

Usually I will not write out all of the details of a calculation like this. I will just write:

Let . Then .

Proof of chain rule: Suppose is differentiable at and is differentiable at
. Then

Hence the theorem will follow from (11.33), the definition of derivative, and the product rule for limits of functions, if we can show that

Since is differentiable at , it follows from lemma 11.17 that

Let be a generic sequence in , such that . Then by (11.34), we have

Since is differentiable at , we have

From this and (11.35) it follows that

Since this holds for a generic sequence in , we have

which is what we wanted to prove. To complete the proof, I should show that is an interior point of . This turns out to be rather tricky, so I will omit the proof.

**Remark**: Our proof of the chain rule is not valid in all cases,
but it
is valid in all cases where you are likely to use it.
The proof fails in the case where every interval
contains a point for which
. (You should check the proof to see where this
assumption was made.) Constant functions satisfy this condition, but if is
constant then is also constant so the chain rule holds trivially in
this case. Since the proof in the general case is more technical than
illuminating, I am going to omit it. Can you find a non-constant function
for which the proof fails?

Also

i.e.,

I will use this relation frequently.

The derivative of can be found by using the quotient rule and the product rule and the chain rule. I will use a trick here which is frequently useful. I have

Now differentiate both sides of this equation using (11.37) to get

Multiply both sides of the equation by to get

This formula is not valid at points where , because we took logarithms in the calculation. Thus is differentiable at , but our formula for is not defined when .

The process of calculating by first taking the logarithm of the
absolute value of
and then differentiating the result, is called *logarithmic
differentiation*.

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