Prove that we can switch $dt$ and $dx$ in the second derivative of a parametric using the limit definition

Question

1. Definitions for this post:

1. The second derivative of a parametric is $\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$ (given by my textbook).
2. Given a parametric of the form $y=f(t),\,x=g(t)$, the latter equation can be algebraically manipulated to yield a new equation $t=h(x)$ such that $y$ can be defined directly in terms of $x$ as $y=f(t)=f(h(x))$.
3. Therefore, the first derivative of a parametric can be given by chain rule: $\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}$.

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2. Premise for the problem:

In trying to derive the second derivative, I start by applying the derivative function to the first derivative: $\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{dy}{dt}\right)\frac{dt}{dx}$ by chain rule, and this expression can be rewritten further as $\frac{\frac{d}{dx}\left(\frac{dy}{dt}\right)}{\frac{dx}{dt}}$. What I don't understand is why you can just switch the $dx$ and $dt$ in the numerator. If we treat $\frac{dy}{dt}$ and $\frac{d}{dx}$ as fractions, I have no problems with this. But given that $\frac{d}{dx}$ is an operator and shorthand for a limit fraction, it's only reasonable that one can prove that $\frac{d}{dx}\left(\frac{dy}{dt}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)$ using the limit definition.

In theory, then, since $f''(x)=\lim_{h\rightarrow0}\frac{f'(x+h)-f'(x)}{h}=\lim_{h\rightarrow0}\frac{\frac{f(x+h+h)-f(x+h)}{h}-\frac{f(x+h)-f(x)}{h}}{h}=\lim_{h\rightarrow0}\frac{\frac{f(x+h+h)-f(x+h)-f(x+h)-f(x)}{h}}{h}=\frac{f(x+h+h)-f(x+h)-f(x+h)-f(x)}{h\cdot h}$, when the second derivative is with respect to a different variable, I should be able to replace the second $h$ with the second differential, right?

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3. The problem:

It would follow, then, that if $\frac{d}{dx}\left(\frac{dy}{dt}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)$,

$$\lim_{\Delta x\rightarrow0}\lim_{\Delta t\rightarrow0}\left[\frac{y(t+\Delta x+\Delta t)-y(t+\Delta x)-y(t+\Delta t)+y(t)}{\Delta x\Delta t}\right]=\lim_{\Delta x\rightarrow0}\lim_{\Delta t\rightarrow 0}\left[\frac{y(x+\Delta t+\Delta x)-y(x+\Delta t)-y(x+\Delta x)+y(x)}{\Delta t\Delta x}\right]$$

...And that's where I'm stuck. Obviously the denominators cancel out, but where do I go from there? If I directly apply the limits at this point I just end up with $0=0$ which isn't necessarily helpful; just because two functions' limits are equal doesn't mean that the functions are equal.

Answer 1

Your use of chain rule in the premise is still problematic. If you mean to be substituting for $\frac{dy}{dx}$ within the outer derivative, then everything needs to stay within the outer derivative, and you get what @nmasanta commented. If you mean to be using the chain rule on the outer derivative, then $\frac{dy}{dx}$ needs to remain intact, and you should get $\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot\frac{dt}{dx}$