# Thread: How do derivate in mathematica 7.0

1. ## How do derivate in mathematica 7.0

I've been able to derivate some function but not the following one:
Code:
Dirac := (m dx[t])/(2 Sqrt[dx[t]^2 + dy[t]^2 + dz[t]^2])
.
I want to derivate the Dirac's function with respect to t. I tried
Code:
D[Dirac[dx, dy, dz], t]
but then I get
Code:
((m
\!$$\*SuperscriptBox["dx", "\[Prime]", MultilineFunction->None]$$[t])/(
2 Sqrt[dx[t]^2 + dy[t]^2 + dz[t]^2]) - (m dx[t] (2 dx[t]
\!$$\*SuperscriptBox["dx", "\[Prime]", MultilineFunction->None]$$[t] + 2 dy[t]
\!$$\*SuperscriptBox["dy", "\[Prime]", MultilineFunction->None]$$[t] + 2 dz[t]
\!$$\*SuperscriptBox["dz", "\[Prime]", MultilineFunction->None]$$[t]))/(
4 (dx[t]^2 + dy[t]^2 + dz[t]^2)^(3/2)))[dx, dy, dz]
which isn't what I'm looking for, I believe (I don't even know what this output mean). Notice that dx, dy and dz aren't the differentials x, y and z. Rather, they're functions which depends on t.
Any idea?

2. Originally Posted by arbolis
I've been able to derivate some function but not the following one:
Code:
Dirac := (m dx[t])/(2 Sqrt[dx[t]^2 + dy[t]^2 + dz[t]^2])
.
I want to derivate the Dirac's function with respect to t. I tried
Code:
D[Dirac[dx, dy, dz], t]
but then I get
Code:
((m
\!$$\*SuperscriptBox["dx", "\[Prime]", MultilineFunction->None]$$[t])/(
2 Sqrt[dx[t]^2 + dy[t]^2 + dz[t]^2]) - (m dx[t] (2 dx[t]
\!$$\*SuperscriptBox["dx", "\[Prime]", MultilineFunction->None]$$[t] + 2 dy[t]
\!$$\*SuperscriptBox["dy", "\[Prime]", MultilineFunction->None]$$[t] + 2 dz[t]
\!$$\*SuperscriptBox["dz", "\[Prime]", MultilineFunction->None]$$[t]))/(
4 (dx[t]^2 + dy[t]^2 + dz[t]^2)^(3/2)))[dx, dy, dz]
which isn't what I'm looking for, I believe (I don't even know what this output mean). Notice that dx, dy and dz aren't the differentials x, y and z. Rather, they're functions which depends on t.

Any idea?
They are not even functions, at best and with some caveats they are functionals.

CB

3. Originally Posted by CaptainBlack
They are not even functions, at best and with some caveats they are functionals.

CB
Ok I see. It makes sense since what I call the Dirac's function arises from Euler-Lagrange equation.
Written differently, I want to derivate the following expression with respect to time: $\frac{m \dot x}{2 \sqrt {\dot x^2 + \dot y^2 + \dot z^2}}$ where the dot represent the derivative with respect to time.
The variables are $\dot x$, $\dot y$ and $\dot z$ all depends on time. Do you think Mathematica can do it?

4. Originally Posted by arbolis
Ok I see. It makes sense since what I call the Dirac's function arises from Euler-Lagrange equation.
Written differently, I want to derivate the following expression with respect to time: $\frac{m \dot x}{2 \sqrt {\dot x^2 + \dot y^2 + \dot z^2}}$ where the dot represent the derivative with respect to time.
The variables are $\dot x$, $\dot y$ and $\dot z$ all depends on time. Do you think Mathematica can do it?
I'm pretty sure Mathematica can, but can't tell you how, I gave up my Mathematica licence a number of years ago now.

CB

5. Ok no problem. I'll be waiting now for any further help from anyone.

6. Try this:

Code:
D[m x'[t]/(2 Sqrt[x'[t]^2 + y'[t]^2 + z'[t]^2]), t]