Hi,
the problem does is not that maxima can not take a partial derivative, but that how you expressed the dependencies:
depends(V, x)$
depends(x, t)$
Due to the chain like manner in which the variables dependencies are expressed, it comes as no surprise that the derivitave calculated used the chain rule:
(%i3) diff(V,x);
dV
(%o3) --
dx
(%i4) diff(V,t);
dx dV
(%o4) -- --
dt dx
If you think about it, this is the correct result even if the d operator is the partial derivative!!! Thus, by letting x be a function of t, %o4 is the correct result for the partial derivative.
The solution is to express the depencies such that the independent variables stay independent of each other, when the normal derivative is then taken, the variables which are not functions of each other appear as constants and hence fall away in the calculation.
The solution is thus:
(%i1) kill(all);
depends(V, [x, t])$
diff(V, x);
diff(V, t);
(%o0) done
(%i1)
(%i2)
dV
(%o2) --
dx
(%i3)
dV
(%o3) --
dt
Vrywaringsklousule / Disclaimer: http://www.nwu.ac.za/it/gov-man/disclaimer.html