Prove the Chain Rule.

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Lesson Parent: 
Differentiation IV: Differentiation Formulas Everyone Must Know
Problem: 

Prove that if $g(x)$ is differentiable at $a$ and $f(x)$ is differentiable at $g(a)$ then $(f\circ g)'(a)=f'(g(a))\,g'(a)$.

SAMPLE SOLUTION

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Solution: 

It is important to note that this proof is more difficult than most of the ones we come across in a basic Calculus course. If you do not have to learn this proof then don't try. Make sure you know what the formula means and can use it when needed.

If you still want to learn the proof then you will also need to learn the following proofs that are used below:

By the definition of the derivative,

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which we can rewrite as

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to get

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Now assume $g(x)$ was such that there was a <Sign in to see all the formulas> where for all <Sign in to see all the formulas> then <Sign in to see all the formulas>. In this case we could write

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because <Sign in to see all the formulas> and we are guaranteed not to divide by zero.

If the limit

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existed (and equaled $L$) then we could write

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since $g(x)$ is differentiable. Now because $g(x)$ is differentiable it is continuous and we can therefore prove that

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Therefore,

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when <Sign in to see all the formulas> around $a$.

Now what happens when our assumption that <Sign in to see all the formulas> for <Sign in to see all the formulas> is not true. This means for all <Sign in to see all the formulas> there exists an $x$ in the interval <Sign in to see all the formulas> such that <Sign in to see all the formulas>. In this case we know that

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But since <Sign in to see all the formulas> in this situation and because $f(x)$ is differentiable at $a$ (i.e. <Sign in to see all the formulas> is a real number) , then we know that <Sign in to see all the formulas>.

Therefore, we can say in general that

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