The chain rule for Calculus in its expanded form

Question

I am currently experiencing some difficulty with the chain rule and there is a debate in my introductory class with a fellow student about the correct way to express it in its expanded form.

Answer 1

I believe you have the right idea. Let me try to give an explanation that will hopefully help out.

The chain rule is much clearer when written in Leibniz notation. Consider a function $f$ which depends on another function $g$, which itself depends on a variable $x$. (i.e. we have $f(g)$ and $g(x)$). Then the chain rule states:

$$ \frac{df}{dx} = \frac{df}{dg} \frac{dg}{dx} $$

In this form you can clearly see that the derivative of $f$ is taken with respect to $g$. It is not taken with respect to $x$. If you knew the derivative of $f$ with respect to $x$ then you wouldn't need the chain rule in the first place.

EDIT: Some clarification based off of comments. The distinction between $\frac{df}{dg}$ and $\frac{df}{dx}$ is more than just semantics. These are completely different things. Often the chain rule is written as:

$$ \frac{d}{dx}f(g(x))=f'(g(x))g'(x) $$

The understanding is that when you place the prime it means 'derivative with respect to the argument.' This causes confusion sometimes because this notation is not explicit about what variable the derivative is in respect to. Note, however, that it is impossible for $f'(g(x))$ to mean $\frac{dg}{dx}$ in the context of the chain rule, because $\frac{dg}{dx}$ is precisely what you are trying to find. It is the element on the left hand side of the equation. If $f'(g(x))$ is taken to mean $\frac{df}{dx}$ then the chain rule says $\frac{df}{dx}=\frac{df}{dx} \frac{dg}{dx}$, implying that $\frac{dg}{dx}=1$. This issue is clearly not just a semantical difference. It produces the wrong result.

Answer 2

The chain rule states that if $h = f \circ g$ then \begin{equation} \tag{$\spadesuit$} h'(x) = f'(g(x)) g'(x) \end{equation} (assuming $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$).

Note that the number $g(x)$ is plugged into the function $f'$, and then we multiply by $g'(x)$.