Comment #2 on issue 2476 by asmeurer: nth order Derivative
http://code.google.com/p/sympy/issues/detail?id=2476
The syntax would have to be different from the non-symbolic case, because
D(x**3, x, a) already means the derivative of x**3 with respect to x and
then with respect to a. I was thinking you could do Derivative(f(x), (x,
n)) to mean the nth derivative of f(x) with respect to x. It should be
backwards compatibile, since we presently don't allow tuples at all. To be
consistant, we should also allow diff(f(x), (x, 3)), but if the arguments
are not nested and symbolic, like diff(f(x), x, n), it must assume it means
the derivative with respect to both x and n as symbols.
By the way, Derivative should internally represent Derivative(f(x), x, 10)
using (x, 10), not (x, x, x, x, x, x, x, x, x, x).
You bring up an interesting point about subs, though. I suppose the only
way to make sense of that is to only let subs change the expresion, not the
variables. In other words, Derivative(x**3, x, a).subs(a, 3) should just
return Derivative(x**3, x, a) because x**3 does not contain a. This also
makes sense since the symbols part of Derivative can only contain symbols
or nonnegative integers, not arbitrary expressions.
And of course, the actual order should allow arbitrary expressions, like
diff(f(x), (x, 2*n)), though if at any point the expression is a number
that is not a nonnegative integer it should raise ValueError (unless we
want to support fractional derivatives).
Finally, I want to note that it might be fun to implement some heuristics
that let us do stuff like
exp(x).diff((x, n))
exp(x)
(x**n).diff((x, n))
factorial(n)
cos(x).diff((x, n))
Piecewise((cos(x), Eq(n%4, 0)), (-sin(x), Eq(n%4, 1)), (-cos(x), Eq(n%4,
2)), (sin(x), Eq(n%4, 3)))
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