Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
3  
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 1

up vote 2 down vote accepted

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.