# A question about the composite function of a derivative

This may seem dumb, but, I'm trying to understand the proof of the chain rule, but here is my issue:

By definition, the derivative is the following:

$f'(a)=\lim\limits_{x\rightarrow a}(\frac{f(x)-f(a)}{x-a})$

So far, so good.

But then, if I were to do a composite function, I would do it like this:

$f'(g(a))=\lim\limits_{x\rightarrow g(a)}(\frac{f(x)-f(g(a))}{x-g(a)})$

I mean, isn't $a$ the input to the $f'(x)$ function?

But the proof states that the composite function is the following:

$f'(g(a))=\lim\limits_{x\rightarrow a}(\frac{f(g(x))-f(g(a))}{x-a})$

And I don't understand why.

-
I think this a notational problem. It's not uncommon with chain rule. When your text wrote $f'(g(a))$, they really meant $(f\circ g)'(a)$, that is the derivative of the composition $f\circ g$ at the point $a$. With that said we have $(f\circ g)'(a) = \lim_{x\rightarrow a}\frac{f(g(x)) - f(g(a))}{x-a}$, which is what they have. Often people can be sloppy with notation (this is a prime example of that) and it can cause a lot more confusion than is necessary. – Cameron Williams Jun 24 '13 at 17:32
Actually, the text was right. I just wrote it like that because I thought it was the same. But now that you point it out, it seems rather obvious that it's a completely different thing! Thanks! – Zequez Jun 24 '13 at 17:42

## 1 Answer

You are correct. Note that: $$f'(g(a))=\lim\limits_{y\rightarrow g(a)}\frac{f(y)-f(g(a))}{y-g(a)}$$ while on the other hand: $$(f \circ g)'(a)=\lim\limits_{x\rightarrow a}\frac{f(g(x))-f(g(a))}{x-a}$$ (To see this, consider the function $h(x)=f(g(x))$.) Proving Chain Rule involves proving that: $$(f \circ g)'(a) = f'(g(a))\cdot g'(a)$$

-
Thanks, it was that, I was just reading $(f \circ g)'(a)$ as if it were the same as $f'(g(a))$, but now I see it's a completely different thing! Thanks! – Zequez Jun 24 '13 at 17:44