How can the chain rule be explained more rigorously? The 'proof' for the chain rule that is often used at school is unsatisfying for me because it treats derivatives as fractions:
$$
\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}
$$
However, the more rigorous proofs that are used in University are unfathomable to me because they are intended for people with a much greater level of background knowledge. Is there a way I can think of the chain rule (perhaps not a rigorous proof) that acknowledges that derivatives are a shorthand for limit expressions, but does not use esoteric notation or complicated methods?
For reference, here is a list of things that I do and don't know:


*

*I know how to differentiate from first principles using $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$

*Apart from the chain rule, I know the product rule and the quotient rule (but again, I don't know the proofs for these rules)

*I know some limit laws (e.g. the quotient law for limits)

*I don't have a rigorous  understanding of limits, but I think I have a good intuitive grasp of them

*Similarly, I have an intuitive understanding of continuous vs. discontinuous functions (continuous = not lifting your pen off the page), but I have not been taught the formal definition for continuity


Thank you for reading.

The chain rule helps us differentiate the composition of functions. It is often the case that the variables $y$ and $x$ can be linked by an intermediate variable $u$. If $y=f(u)$, where $u=g(x)$, then $y=f(g(x))$, meaning that $y$ and $x$ are linked through the composite function $f \circ g$. Here is an informal argument, using differentials, that
$$
\frac{dy}{dx}=f'(g(x))g'(x) \, .
$$
If $y=f(g(x))$, then by definition
$$
dy=f(g(x+dx))-f(g(x)) \, .
$$
To make things a little easier to look at, let $u=g(x)$ and $du=g(x+dx)-g(x)$. Then we have
\begin{align}
dy &= f(u+du)-f(u) \\
&= f(u)+f'(u)du - f(u)\\
&= f'(u)du \\
\end{align}
and so
$$
\frac{dy}{dx}=f'(u)\frac{du}{dx}=f'(g(x))g'(x) \, .
$$
