Proof of Chain Rule and Intuition about what it Does I looked up a proof of the chain rule here https://web.williams.edu/Mathematics/lg5/A37W12/Chain.pdf

which made sense from a computational standpoint. However, I do not understand what intuitive details about the process of the chain rule or why it works, from this proof.
Let $h(x) = f(g(x))$. Intuitively, $h'(x)$ is a measure of how much $h$ changes per unit $x$ for a small change in $x$:

A small change $\delta x$ in $x$ causes the change $g'(x) \delta x$ in the value of g(x). The change in the value of $h(x) = f(g(x))$ as a result of the change in $g(x)$ is then given by $f'(g(x))$ multiplied by the change in $g(x)$ which was $g'(x) \delta x$. So the total change in $h$ is given by $f'(g(x))g'(x) \delta x$.

Divide both sides by $\delta x$ and take the limit of $\delta x \to 0$ (where the approximations become exact) and you have your chain rule: $h'(x) = f'(g(x)) g'(x)$.
Here is a proof of the chain rule. If $h(x)=f(g(x))$ then:
$$h(x + Δx) = f(g(x + Δx))$$
The gradient equation is of course:
$$f(x + Δx) = f(x) + f'(x)Δx$$
where $f(x)$ is any smooth function. Therefore we can say:
\begin{align*}
h(x + Δx) &amp;= f[g(x) + g'(x)Δx]\\
&amp;=f(g(x)) + f'(g(x))g'(x)Δx
\end{align*}
This is a direct consequence of the gradient equation i.e. it applies to both finite differences and calculus itself without any other treatment e.g. taking the standard part (or limit) or neglecting infinitesimals (which is unusual). Consequently $Δx→dx$, so substituting and rearranging:
$$\frac{h(x + dx) - h(x)}{dx} = f'(g(x))g'(x)$$
Therefore, the chain rule is:
$$h'(x) = f'(g(x))g'(x)$$
The above functional form can be directly converted to the differential form:
$$\frac{dh}{dx} = \frac{df}{dg}\frac{dg}{dx}$$
Note that the first numerators on the two sides are different. Usually they are written the same, but that may be a case of a bad convention. However, the equation still balances - $df$ is the differential as a function of $g$. This is implied by the denominator $dg$ so it doesn't have to be stated. If we want to express the LHS in terms of $f$ it should be:
$$\frac{d(f(g))}{dx} = \frac{df}{dg}\frac{dg}{dx}$$

EDIT: This was the first answer.