A question about the composite function of a derivative

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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.

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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: $$ (f \circ g)'(a) = f'(g(a))\cdot g'(a) $$