Backwards chain rule

[needmathhelp]

if somebody can find a counter example that would be excellent. More excellent if he can prove it.


Let f be a differenciable function at c and let g be a function which is defined on an open interval containing f(c) such that (g o f) is differenciable at c. Show that g is differenciable at c.

thanks a lot.

[elgiovo]

Ok, you're right. I have to look conditional sentences over. So, what does Spivak say about that?

[Sandokan2]

"elgiovo":So if $g$ wouldn't be differentiable at $f(c)$ then $(g@f)'(c)$ wouldn't exist

How do you know that?

This is precisely what we are supposed to prove.

Please forgive me, sir, if I dare make a little remark of a grammatical nature

The correct sentence reads as follows:

if $g$ weren't differentiable

I am however your humble servant

[elgiovo]

"Sandokan":[quote="elgiovo"]By hypothesis, $f$ and $g @ f$ are differentiable at $c$. The derivative of $g @ f $ at $c$ is $f'(c)cdotg'(f(c))$, by chain rule.

I beg your pardon, sir, but you are mistaken here. Since we don't know that $g$ is differentiable, the chain rule doesn't apply.[/quote]
We know that $g @ f$ is differentiable at $c$, so such a derivative exists. If $g$ was differentiable, then it would be $f'(c)cdotg'(f(c))$. And that is ok. So if $g$ wouldn't be differentiable at $f(c)$ then $(g@f)'(c)$ wouldn't exist, but we know it to exist, so $g$ has to be differentiable.

[Thomas16]

ok but I guess elgiovo is right when saying that we have to prove that g is differentiable at f(c) and not at c...

[Sandokan2]

"elgiovo":By hypothesis, $f$ and $g @ f$ are differentiable at $c$. The derivative of $g @ f $ at $c$ is $f'(c)cdotg'(f(c))$, by chain rule.

I beg your pardon, sir, but you are mistaken here. Since we don't know that $g$ is differentiable, the chain rule doesn't apply.

[elgiovo]

By hypothesis, $f$ and $g @ f$ are differentiable at $c$. The derivative of $g @ f $ at $c$ is $f'(c)cdotg'(f(c))$, by chain rule. Always by hypothesis, both $f'(c)$ and $g'(f(c))$ have to exist for $(g @ f)$ to exist, so $g$ is differentiable at $f(c)$. If you wanted to prove that $g$ is differentiable at $bbc$, that is a strange problem, because you have to suppose that $f$ has got a fixed point at $c$, but I believe you have made a mistake in copying.

[Sandokan2]

Gentlemen

this is rather an interesting question

I seem to remember having seen it in 'Calculus on Manifolds' by Spivak

anyway... I won't conceal the fact that English is not my favourite language