> assumes that all symbols that do not have an explicit dependence on the variable of differentiation are constants
`idiff` will allow you to do the differentiation of symbols without functions, e.g. `dydx for idiff(2*x - y**2, y, x) -> 1/y` /c On Wednesday, April 20, 2022 at 7:37:33 AM UTC-5 [email protected] wrote: > On Apr 19, 2022, at 10:48 PM, Andre Bolle <[email protected]> wrote: > > Here's what I did. You will notice that the second derivative couldn't see > the 'x', which had been replaced by psi. > > Yes, that is a “feature” of sympy, which assumes that all symbols that do > not have an explicit dependence on the variable of differentiation are > constants. Most of the time that works well. You can get around that by > specifying that psi is a function of x. See the sympy documentation on > functions. I have been fiddling with how to make it take more general > derivatives, but need to figure out a definition of an infinitesimal that > will function consistently within the sympy environment. > > > I do like the equation annotation. (eq1), (eq2), (eq3), etc. Nice. > > Glad you like it. I sometimes collapse the code blocks. This then leaves > you with the results of each step as might be provided in a traditional > derivation. You can also pretty it up with comments in markdown cells > between the each step. > > Jonathan > -- You received this message because you are subscribed to the Google Groups "sympy" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/b341e69d-5a59-4df8-805f-926bc7c140c9n%40googlegroups.com.
