In Maple, I have a function $f(x(t),y(t),t)$ that I want to differentiate with respect to $t$. I know the command for partial derivative $\frac{\partial f}{\partial x}$,$\frac{\partial f}{\partial y}$,$\frac{\partial f}{\partial t}$, etc in Maple. But I can't find a way to find total derivative $\frac{df}{dt}$.

New contributor
ilawid is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
  • 2
    $\begingroup$ So just do it explicitly $\endgroup$ Nov 24 at 15:54
  • $\begingroup$ Have you tried the diff from the Physics package? $\endgroup$
    – Tyberius
    Nov 25 at 2:13

You have this function:

$$ f = f(x,y,t) $$


$$ df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial t} dt $$


$$ \frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial t} $$

You said that, you are able to calculate $\frac{\partial f}{\partial x}$, $\frac{\partial f}{\partial y}$, and $\frac{\partial f}{\partial t}$. So, just calculate them and I assume you know the relations of $x = x(t)$, $y = y(t)$ as well. So, everything is ready to find the $\frac{df}{dt}$ here.

  • $\begingroup$ Thanks, but I need a general symbol for this in MAPLE. I need to find total derivatives several times. Doing the manual process is not feasible. $\endgroup$
    – ilawid
    Nov 24 at 20:25
  • 1
    $\begingroup$ @ilawid I worked with Maple many years ago, but I believe it should be easy to automate this process. Basically it's a function that you put your symbolical $f$, $x=x(t)$, and $y=y(t)$ into it and it just calculates $\partial_{x} f$, $\partial_{y} f$, $\partial_{t} f$, $\dot{x}$, and $\dot{y}$ and returns $\dot{f}$ based on that formula. I'm not sure why you think you need to do it manually. Note: at the end $\dot{f}$ might not be something really compact or even it's not guaranteed that you would be able to write it down based on only $t$ parameter. It just depends on explicit formula of $f$. $\endgroup$ Nov 24 at 20:50

Your Answer

ilawid is a new contributor. Be nice, and check out our Code of Conduct.

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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