Derivative existence theorem Has anyone here heard of the Derivative existence theorem? 

Derivative existence theorem: For $f$ defined on some interval
  including $a$, $f$ is differentiable at $a$ if and only if there
  exists a function $\varphi $, which is continuous at $a$, and
  satisfies $\forall x\leq $ interval, $f(x) - f(a) = \varphi(x)(x-a)$. 
In this case, $f'(a)=\varphi(a)$.

I heard that the Chain Rule Theorem for derivatives is much easier to prove using the existence theorem.
 A: I hadn't heard of the Derivative Existence Theorem, but the statement looks familiar. Below is the proof as well as the proof of the Chain Rule using it.
Proof of the Derivative Existence Theorem: Let $f$ be a function defined on $(a - \varepsilon, a + \varepsilon)$ for some $\varepsilon > 0$ which is differentiable at $a$. Define 
$$\varphi(x) = \begin{cases}
\dfrac{f(x) - f(a)}{x-a} & \text{if}\ x \neq a\\
&\\
\lim\limits_{x\to a}\dfrac{f(x) - f(a)}{x-a} & x = a.
\end{cases}$$
Then $\varphi(x)$ is continuous at $a$, $f(x) - f(a) = \varphi(x)(x - a)$, and $\varphi(a) = f'(a)$.
Conversely, if $f(x) - f(a) = \varphi(x)(x - a)$ for some $\varphi$ which is continuous at $a$, then for $x \neq a$
$$\varphi(x) = \frac{f(x) - f(a)}{x - a}.$$
As $\varphi$ is continuous at $a$,
$$\varphi(a) = \lim_{x \to a}\varphi(x) = \lim_{x \to a}\frac{f(x) - f(a)}{x - a}.$$
So $f'(a)$ exists and is equal to $\varphi(a)$. 
$\hspace{16.8cm}\square$

Chain Rule: Let $g$ be a function which is differentiable at $a$ and let $f$ be a differentiable at $g(a)$. Then $f\circ g$ is differentiable at $a$ and $(f\circ g)'(a) = f'(g(a))g'(a)$.
Proof of the Chain Rule: We have $g(x) - g(a) = \varphi_g(x)(x - a)$, and $f(x) - f(g(a)) = \varphi_f(x)(x - g(a))$ where $\varphi_g$ is continuous at $a$ with $g'(a) = \varphi(a)$, and $\varphi_f$ is continuous at $g(a)$ with $f'(g(a)) = \varphi_f(g(a))$. Then 
$$(f\circ g)(x) - (f\circ g)(a) = f(g(x)) - f(g(a)) = \varphi_f(g(x))(g(x) - g(a)) = \varphi_f(g(x))\varphi_g(x)(x-a).$$
As $g$ is differentiable at $a$, $g$ is continuous at $a$ so $\varphi_f(g(x))$ is continuous at $a$, and therefore $\varphi_f(g(x))\varphi_g(x)$ is continuous at $a$. By the Derivative Existence Theorem, $f\circ g$ is differentiable at $a$ and $(f\circ g)'(a) = \varphi_f(g(a))\varphi_g(a) = f'(g(a))g'(a)$. 
$\hspace{16.8cm} \square$
