Sources: Calculus: Early Transcendentals (6 edn 2007). p. 206, Section 3.4. Question 95.
Elementary Differential Equations (9 edn 2008). p. 165, Section 3.3, Question 34.
Quandary: If $y=f(x)$, and $x = u(t)$ is a new independent variable, where $f$ and $u$ are twice differentiable functions, what's $\dfrac{d^{2}y}{\mathrm{d}t^{2}} $?
Answer: By the chain rule, $\dfrac{\mathrm{d}y}{\mathrm{d}t} = \dfrac{\mathrm{d}y}{\mathrm{d}x} \dfrac{\mathrm{d}x}{\mathrm{d}t} $. Then by the product rule,
$\begin{align} \dfrac{d^{2}y}{\mathrm{d}t^{2}} & \stackrel{\mathrm{defn}}{=}\dfrac{d}{\mathrm{d}t} \dfrac{\mathrm{d}y}{\mathrm{d}t} \\ & = \color{forestgreen}{ \dfrac{d}{\mathrm{d}t} \dfrac{\mathrm{d}y}{\mathrm{d}x} } & \dfrac{\mathrm{d}x}{\mathrm{d}t} & +\frac{\mathrm{d}y}{\mathrm{d}x} \quad \dfrac{d}{\mathrm{d}t} \dfrac{\mathrm{d}x}{\mathrm{d}t} \\ & = \color{red}{\frac{d^{2}y}{\mathrm{d}x^{2}}(\frac{\mathrm{d}x}{\mathrm{d}t})} & \dfrac{\mathrm{d}x}{\mathrm{d}t} & + \frac{\mathrm{d}y}{\mathrm{d}x} \quad \frac{d^{2}x}{\mathrm{d}t^{2}} \\ & = \frac{d^{2}y}{\mathrm{d}x^{2}}(\frac{\mathrm{d}x}{\mathrm{d}t})^{2} & &+\frac{\mathrm{d}y}{\mathrm{d}x} \quad \frac{d^{2}x}{\mathrm{d}t^{2}} \end{align} $
Why $\color{forestgreen}{ \dfrac{d}{\mathrm{d}t} \dfrac{\mathrm{d}y}{\mathrm{d}x} } = \color{red}{\dfrac{d^{2}y}{\mathrm{d}x^{2}}(\dfrac{\mathrm{d}x}{\mathrm{d}t})} $? Please explain informlly and intuitively.
Think of $\frac{dy}{dx}$ as a function of $x$; say it is $f(x)$. Then $x$ is also a function of $t$. So, by the exact same chain rule argument that you gave, $$ \frac{d}{dt}\left[\frac{dy}{dx}\right]=\frac{d}{dt}[f(x)]=\frac{df}{dx}\cdot\frac{dx}{dt}. $$ But $$ \frac{df}{dx}=\frac{d}{dx}\left[\frac{dy}{dx}\right]. $$