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Given $$ \left\{ \begin{align*} x &= f(t)\\ y &= g(t) \end{align*}\right. $$

We can compute $\frac{dy}{dx}$ simply by $$ \frac{dy}{dx}=\frac{g'(t)}{f '(t)} $$

However when I tried to compute $\frac{d^2y}{dx^2}$, I met some problem. I've tried the chain rule but it seemed failed.

Can you please help? Thank you.

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    $\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)$, and $\frac{d}{dx}(\cdots)=\frac{dt}{dx}\frac{d}{dt}(\cdots)$2011-11-07
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    Are you sure that the derivatives you want are really $\frac{d^ny}{dx^n}$ and not $\frac{d^n(x,y)}{dt^n}$?2011-11-07
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    @HenningMakholm: Yes.2011-11-07
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    [Parametric derivatives](http://calculusapplets.com/parametric.html)2011-11-07
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    This question is similar to [this question](http://math.stackexchange.com/q/68988/) and [this question](http://math.stackexchange.com/q/49734/). Do you have a question not answered by one of their answers?2011-11-07
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    $$\frac{d^2y}{dx^2}=\frac{\dot{x}\ddot{y}-\dot{y}\ddot{x}}{\dot{x}^{3}}$$ , where $\dot{x}=x'_t$ and $\ddot{x}=x''_t$2011-11-07
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    @robjohn: Ah..I'm very sorry that I forgot to search before asking after a long time not using this site... Thank you for the link.2011-11-07

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Just set $y'={dy\over dx}$, then $${d\over dt} y'={dy'\over dx}{dx\over dt};$$ whence $${d^2y\over dx^2}={ {dy' / dt} \over {dx/dt}}.$$

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    Thank you for your answer. I just tried to apply '$\frac{d}{dx}$' and did not see that applying '$\frac{d}{dt}$' first then we can solve '$\frac{d}{dx}$' out.2011-11-07