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If $\frac{dy}{dt}$ and $\frac{dx}{dt}$ exist, then does $\frac{dy}{dx}$ always exist when $\frac{dx}{dt} \not=0$?

Indeed, this is a very simple question. Sorry but I'm just a beginner for Calculus so I don't know that much in depth. But here is the reason for asking.

During my studying the parametric curves, we learn that

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

given that $\frac{dx}{dt} \not=0$. But from the Chain Rule, when $\frac{dy}{dt}$ and $\frac{dt}{dx}$ exist,

$$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}$$

But as far as I know, for the given information, we cannot say that even though $\frac{dy}{dt}$ and $\frac{dt}{dx}$ exist, $\frac{dy}{dx}$ might not be the same as $\frac{dy/dt}{dx/dt}$ since

$$\frac{dt}{dx}=\frac{1}{\frac{dx}{dt}}$$

if the inverse of $t(x)$ exists (as far as I know... I'm not sure). So for my short knowledge, I'm guessing there might be a case even though $\frac{dy}{dt}$ and $\frac{dx}{dt}$ exist, $\frac{dy}{dx}$ might not exist(or have different value comapared to the suggested one up there). That might happen when $t$ is not one-to-one function, thus having no inverse function against $x$!

Thanks.

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Counter example: $$x=\sin t$$ $$y=\cos t$$ $$\frac {dy}{dx}=-\tan t$$ $\frac {dy}{dx}$ does not exist when t=$\frac {\pi}{2}$ but $\frac{dx}{dt}$ exist at x=$\pi/2$

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  • $\begingroup$ So much thank!! $\endgroup$
    – Taxxi
    Nov 19, 2013 at 19:13

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