# How to the take derivatives of a parametric function? [closed]

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?

-

## closed as off-topic by Artes, Kuba, rasher, Yves Klett, ubpdqnApr 24 at 10:58

• The question does not concern the technical computing software Mathematica by Wolfram Research. Please see the help center to find out about the topics that can be asked here.
If this question can be reworded to fit the rules in the help center, please edit the question.

Is this a question about the software Mathematica or about mathematics? If mathematics, then try math.stackexchange.com. –  Michael E2 Apr 24 at 1:46
basically you use y'[x]=y'[t]/x'[t] –  belisarius Apr 24 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 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 at 4:23
@Artes. It seems clear enough to me. I'm answering it. –  m_goldberg Apr 24 at 4:50

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.

-