How to think about this simple differential calculus expression?

Question

In the course of doing differential equation substitutions, I often come across the expression $\frac{d}{dt}\frac{dy}{dx}$, where $y$ is a function of $x$, and $x$ itself is a function of $t$. For some reason, I have a hard time figuring out how to expand this. One possibility is to multiply the inside by $\frac{\frac{dx}{dt}}{\frac{dx}{dt}}$, yielding $\frac{d}{dt}\frac{\frac{dy}{dx}\frac{dx}{dt}}{\frac{dx}{dt}} \implies \frac{(\frac{d}{dt}(\frac{dy}{dx}\frac{dx}{dt}))\frac{dx}{dt} - (\frac{dy}{dx}\frac{dx}{dt})\frac{d}{dt}\frac{dx}{dt}}{(\frac{dx}{dt})^2}$, which looks nice because $\frac{dy}{dx}\frac{dx}{dt} = \frac{dy}{dt}$. However, to proceed further I would need to use the expanded version, and then the initial expression $\frac{d}{dt}\frac{dy}{dx}$ reappears. Is there a simple chain rule here I'm not spotting? How should I think about expanding this expression?

Answer 1

Let $y=f(x)$ and let $x(t)$ be a given function. Notice that $\frac{dy}{dx}=f'(x)$, so we need to calculate $\frac{d}{dt}(f'(x(t)).$ Applying the chain rule, we find this is $f''(x(t))\cdot \frac{dx}{dt}$. Translating back to your notation, this means that $\frac{d}{dt} \frac{dy}{dx}=\frac{d^2 y}{dx^2} \frac{dx}{dt}.$