Higher order derivative of parameter curves

Question

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.

$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)$, and $\frac{d}{dx}(\cdots)=\frac{dt}{dx}\frac{d}{dt}(\cdots)$ — pharmine, Nov 07 '11 at 15:50
Are you sure that the derivatives you want are really $\frac{d^ny}{dx^n}$ and not $\frac{d^n(x,y)}{dt^n}$? — hmakholm left over Monica, Nov 07 '11 at 15:53
This question is similar to this question and this question. Do you have a question not answered by one of their answers? — robjohn, Nov 07 '11 at 16:33
$$\frac{d^2y}{dx^2}=\frac{\dot{x}\ddot{y}-\dot{y}\ddot{x}}{\dot{x}^{3}}$$ , where $\dot{x}=x'_t$ and $\ddot{x}=x''_t$ — Pedja, Nov 07 '11 at 16:35
@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. — Roun, Nov 07 '11 at 16:39

score 1 · Accepted Answer · answered Nov 07 '11 at 16:05

1

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

answered Nov 07 '11 at 16:05

David Mitra

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 07 '11 at 16:46

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