I'm trying to follow some calculus lecture notes and I can reproduce an argument for the sum and product rule. Can I make an argument like this for the chain rule in a similar style, using the definition of $df$?
$$ df := f(x+dx) - f(x) $$
Sum rule
$$ f(x) := g(x) + h(x) $$
\begin{align} df &= f(x+dx) - f(x) \\ &= g(x+dx)+h(x+dx)-(g(x)+h(x)) \\ &= dg + g(x) + dh + h(x) - g(x) - h(x) \\ &= dg + dh \end{align} where step (3) uses $dg=g(x+dx)-g(x)$ and likewise for $dh$.
Product rule
$$f(x) := g(x)h(x)$$
\begin{align} df &= f(x+dx) - f(x) \\ &= g(x+dx)h(x+dx) - g(x)h(x) \\ &= (dg + g(x))(dh+h(x)) - g(x)h(x) \\ &= dgdh + dgh + gdh + gh - gh \\ &= dgdh + dgh + gdh \\ &= dgh + gdh \end{align}
using the same rewrite as above.
Chain rule
$$f(x) := g(h(x))$$
\begin{align} df &= f(x+dx)-f(x) \\ &= g(h(x+dx)) - g(h(x)) \\ &= g(dh + h(x)) - g(h(x)) \\ &= \quad ? \end{align}
You could do something like this: (Assuming that the derivative of $f$, $f^\prime$ is defined as $\frac{f(x+dx)-f(x)}{dx}$.) $$ df=f(x+dx)-f(x) \\ =g(h(x+dx))-g(h(x)) \\ =g(h(x)+dh)-g(h(x)) \\ =\frac{g(h(x)+dh)-g(h(x))}{dh}dh \\ =g^\prime(h(x))dh. $$