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I have a parametric curve of the form

$x(t)=2t^3+3t^2-12t,\ y(t)=2t^3+3t^2+1$

How do I take the derivative of that function in Mathematica? Is there a certain function I should use?

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closed as off-topic by Artes, Kuba, ciao, Yves Klett, ubpdqn Apr 24 '14 at 10:58

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Is this a question about the software Mathematica or about mathematics? If mathematics, then try math.stackexchange.com. –  Michael E2 Apr 24 '14 at 1:46
basically you use y'[x]=y'[t]/x'[t] –  belisarius Apr 24 '14 at 2:31
Could you show some Mathematica code that you have already tried? It's not clear what is causing you difficulties. –  Jens Apr 24 '14 at 4:15
I'm closing it since: this question is ill-posed (unclear what function), even if it is well-posed it can be easily found in the documentation. Last but not least the user is unregistered. –  Artes Apr 24 '14 at 4:23
@Artes. It seems clear enough to me. I'm answering it. –  m_goldberg Apr 24 '14 at 4:50

1 Answer 1

The Mathematica built-in function D is what you are looking for.

For your problem it would be used like this:

Clear[f, df]
f[t_] = {2 t^3 + 3 t^2 - 12 t, 2 t^3 + 3 t^2 + 1};
ParametricPlot[f[t], {t, -2.5, 2.3}]


A function you can use as the derivative of f is defined by

df[t_] = D[f[t], t]
{-12 + 6 t + 6 t^2, 6 t + 6 t^2}

For example, here it is used to make a plot.


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