# Chain Rule Fake Proof?

Let $C,D$ be infinite connected sets in $\mathbb{R}$.

Let $f:C\rightarrow \mathbb{R} , g:D\rightarrow \mathbb{R}$ be functions such that $f$ is differentiable at some $x\in C$ and $g$ is differentiable at $f(x)$.

Proof goes here; Define $u(t)=\frac{f(t)-f(x)}{t-x} - f'(x), \forall t\in C\setminus \{x\}$.

Define $v(s)=\frac{g(s)-g(f(x))}{s-f(x)} - g'(f(x)), \forall s\in D\setminus \{f(x)\}$.

Then, $g(f(t))-g(f(x)) = (t-x)(f'(x)+u(t))(g'(f(x)) + v(f(t)))$.

Here, it seems to me the equation is false.

Fix $\epsilon >0$. Since $g$ is differentiable at $f(x)$, $\exists \delta>0$ such that $\forall s\in D\setminus \{f(x)\}, 0<d(s,f(x))<\delta \Rightarrow |\frac{g(s)-g(f(x))}{s-f(x)} - g'(f(x))|<\epsilon$.

Since $f$ is continuous at $x$, $\exists \delta_0 >0$ such that $\forall t\in C\setminus \{x\}, 0<d(t,x)<\delta_0 \Rightarrow d(f(t),f(x))<\delta$.

See, if $\forall \alpha>0, \exists t\in C\setminus \{x\}$ such that $f(t)=f(x)$ and $0<d(t,x)<\alpha$, we cannot pick a positive real $\delta_0$ such that $\forall t\in B(x,\delta_0), v(f(t))$ is defined...

I want a precise proof

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You want a precise proof of what? –  joriki Nov 13 '12 at 19:04
@joriki Proof for chaine rule. By the way, now i proved it precisely. Anyway is it still true that above equality is generally false? (Rudin PMA p.105) –  Katlus Nov 13 '12 at 19:15