Question: How can D((x+y)^a) be calculated as D((x+y)^2 is?

Hello people in mapleprimes,

I hope that you will give me an answer about the calculation surrounding D.
D((x+y)^2) can be calculated with maple, but D((x+y)^a), where a is a constant, cannot at least directly.
And, I have the code, which Mr. Carl Love kindly gave me at
, which is

h:= F-> expand(evalindets(D(F)/F, specfunc(D), d-> op(d)*'h'(op(d))));

This works for (x+y)^2, but does not work for (x+y)^a.
And, I think this reason is that the behavior of D to (x+y)^2 is not the same
 as that to (x+y)^a.
And, I want to modify this code.
But, I can't .
So, I am asking this question. How should I modify the above code so that it works for (x+y)^a.
I'm so sorry for a lot of impoliteness.
I will be very glad if you give me an answer.

Take care.


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