Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)$, 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
up vote 1 down vote accepted

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.