Here's what I'm proving:
Let $g$ be differentiable at $x$ and $f$ be differentiable at $g(x)$. Then, $f \circ g$ is differentiable at $x$ and it is the case that:
$$(f \circ g)'(x) = f'(g(x)) \cdot g'(x)$$
I'm kind of approaching this without using the definition of the limit because I tried using that and it didn't really work very well. So, I want to try another approach.
If $g$ is differentiable at $x$, then $g$ is continuous at $x$. From a previously proven result, it is the case that:
$$\lim_{\Delta x \to 0} \frac{\Delta g}{\Delta x} = g'(x) \iff \exists \alpha(\Delta x): \frac{\Delta g}{\Delta x} = g'(x) + \alpha(\Delta x) \land \lim_{\Delta x \to 0} \alpha = 0$$
Similarly:
$$\lim_{\Delta g \to 0} \frac{f(g(x+\Delta x)) - f(g(x))}{\Delta g} = f'(g(x)) \iff \exists \beta(\Delta g): \frac{\Delta f*}{\Delta g} = f'(g(x)) + \beta(\Delta g) \land \lim_{\Delta g \to 0} \beta = 0$$
Define $\phi(x) = f(g(x))$. Then:
$$\Delta \phi = \phi(x+\Delta x) - \phi(x) = f(g(x+\Delta x)) - f(g(x))$$
$$\Delta \phi = f(g(x) + \Delta g) -f(g(x)) = \Delta f*$$
$$\Delta \phi = f'(g(x)) \Delta g + \beta \Delta g$$
$$\Delta \phi = f'(g(x)) \cdot [g'(x) \Delta x + \alpha \Delta x] + \beta [g'(x) \Delta x + \alpha \Delta x]$$
$$\frac{\Delta \phi}{\Delta x} = f'(g(x))g'(x) + f'(g(x)) \alpha + \beta g'(x) + \beta \cdot \alpha$$
Taking the limit as $\Delta x \to 0$, we have:
$$\phi'(x) = f'(g(x))g'(x) + g'(x) \cdot \lim_{\Delta x \to 0} \beta(\Delta g)$$
Now, we know that since $g$ is continuous, it follows that $\Delta g \to 0$ as $\Delta x \to 0$. So, we define $\beta(0) = 0$ for our convenience and so, $\beta(\Delta g)$ is continuous at $\Delta g = 0$. Hence:
$$\phi'(x) = f'(g(x))g'(x) + g'(x) \cdot 0 = f'(g(x))g'(x)$$
That proves the desired result.
Does the proof above work? If it doesn't, why? How can I fix it?