Chain rule proof verification

Question

Is the following a correct way to prove the chain rule. I am in doubt because of how I am presented the proof in other sources that I think over complicates things (the prior statement might be biased) :

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Chain Rule: If $f(u)$ is a differentiable at point $u=g(x)$, and $g(x)$ is a differentiable at $x$, then $f(g(x))$ is differentiable at $x$, and $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$, where $\frac{dy}{du}$ is evaluated at $u=g(x)$

Proof:

If $g(t)$ is not $g(x)$ for $t$ near $x$ then: $\frac{dy}{dx}=\lim_{t \to x} \frac{f(g(x))-f(g(t))}{g(x)-g(t)} \times \frac{g(x)-g(t)}{x-t} = f'(g(x)) * g'(x)$

If $g(t)=g(x)$ for $t$ near $x$ then $g'(x)$ is $0$ so $f'(g(x))*g'(x)=0$. However, $\frac{dy}{dx}=\lim_{t \to x} \frac{f(g(x))-f(g(t))}{x-t}=0$ because $g(x)=g(t) \implies f(g(x)) = f(g(t))$.

Answer 1

You cannot divide by $g(t_n) - g(x)$ in the general case. Here is a hint : $f$ is differentiable at $x$ if and only if there exists a function $\varepsilon$ with $\varepsilon(h) \to 0$ as $h\to 0$, and a number $a$ (which is $f'(x)$) such that \begin{align} f(x + h) = f(x) + a h + h \varepsilon(h) \end{align}

Write it for $f$ and $g$, and carefully do the composition.

Answer 2

Sorry, but to be honest I really do not understand your proof. In my opinion, the simplest way I feel to prove the Chain Rule is via Caratheodory's theorem, which states that if $f(x)$ is differentiable at $x = c$, then there exists a function $\phi(x)$ such that:

(i) $\phi(x)$ is continuous at $x =c$

(ii) $f(x) - f(c) = \phi(x) (x-c)$

For your example, given that $g(x)$ is differentiable at $x = c$, there exists a function $\alpha(x)$ such that:

(i) $\alpha(x)$ is continuous at $x =c$

(ii) $g(x) - g(c) = \alpha(x) (x-c)$

(iii) It is also clear that $g'(c) = \alpha(c)$

Again, given that $f(x)$ is differentiable at $u = g(c)$, there exists a function $\beta(y)$ such that:

(i) $\beta(y)$ is continuous at $y = u$

(ii) $f(y) - f(u) = \beta(y) (y - u)$

(iii) It is also clear that $f'(u) = \beta(u)$

Now our goal is to show that $h = f \circ g$ is differentiable at $c$. If this is the case, then there SHOULD exist a function $\gamma (x)$ such that:

(i) $\gamma(x)$ is continuous at $x = c$

(ii) $h(x) - f(c) = \gamma(x) (x - c)$

(iii) $h'(c) = \gamma(c)$

Notice that: $$f(g(x)) - f(g(c)) = \beta(g(x))(g(x) - g(c)) = \beta(g(x)) \cdot \alpha(x) (x - c)$$

Hence we can define $\gamma(x) = \beta(g(x)) \cdot \alpha(x)$ which will conclude the proof. By Carotheodory's Theorem, since $\gamma$ exists, then $h = f \circ g$ is differentiable at $x = c$

I must say that I am not rigorous here because I did not mention anything about the intervals the functions were defined on. But I am sure you can do it by simply matching the domain of $\gamma$ to that of $\alpha$