# Higher order derivative of parameter curves

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.

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$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)$, and $\frac{d}{dx}(\cdots)=\frac{dt}{dx}\frac{d}{dt}(\cdots)$ – pharmine Nov 7 '11 at 15:50
Are you sure that the derivatives you want are really $\frac{d^ny}{dx^n}$ and not $\frac{d^n(x,y)}{dt^n}$? – Henning Makholm Nov 7 '11 at 15:53
@HenningMakholm: Yes. – Roun Nov 7 '11 at 15:58
Parametric derivatives – pedja Nov 7 '11 at 16:06
This question is similar to this question and this question. Do you have a question not answered by one of their answers? – robjohn Nov 7 '11 at 16:33

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. – Roun Nov 7 '11 at 16:46