Step from $\Big(\frac{d}{dx}\frac{dy}{dt}\Big)t$ to $\Big(\frac{d^2y}{dt^2}\frac{dt}{dx}\Big)t$

Question

From this page between (2) and (3): https://math.dartmouth.edu/archive/m23s06/public_html/handouts/euler_eqns.pdf

$$\Big(\frac{d}{dx}\frac{dy}{dt}\Big)t=\frac{d^2y}{dt^2}\frac{dt}{dx}t$$

I always had difficulties with the chain rule, especially in Leibniz notation.

Where does the second derivative with respect to $t$ come from?

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PDF transcribed with color for emphasis

$$\begin{align} \frac{d^2 y}{dx^2} &= \frac{d}{dx} \frac{dy}{dx}\\ &=\frac{d}{dx}\left(\frac{dy}{dt} t\right)\\ &=\color{red}{\left(\frac{d}{dx}\frac{dy}{dt}\right)t}+\frac{dy}{dt}\frac{dt}{dx}\\ &=\color{red}{\frac{d^2 y}{dt^2}\frac{dt}{dx}t}+\frac{dy}{dt}\frac{dt}{dx}\\ &=\frac{d^2 y}{dt^2}t^2+\frac{dy}{dt}t \end{align}$$ Where we have again used the fact that $dx/dt = t$

Answer 1

It seems that they (somewhat perversely) have $t$ as a function of $x$. Then the above is really the chain rule: To differentiate $\frac{dy}{dt}$ with respect to $x$, we must first differentiate with respect to $t$ and then $t$ by $x$ to uncover the rate of change in $x$. By the chain rule, these are multiplied $$ \frac{d^2y}{dt^2}\frac{dt}{dx} $$

Answer 2

Perhaps, it'd help to view it as the below operator acting on $y$. Note that the $t$ at the end in your expression in the LHS and RHS is a just a multiplicative factor and not acted upon by the operator.$$ \frac{\mathrm{d}}{\mathrm{d} x} \left( \frac{\mathrm{d}}{\mathrm{d} t}\right)= \left(\frac{\mathrm{d }t}{\mathrm{d} x} \right)\frac{\mathrm{d}}{\mathrm{d} t}\left( \frac{\mathrm{d}}{\mathrm{d} t}\right) = \left(\frac{\mathrm{d }t}{\mathrm{d} x} \right)\frac{\mathrm{d^2}}{\mathrm{d} t^2}$$