| Author |
Comment/Response |
Michelle
|
 09/22/04 3:45pm
Is there a way I can make Mathematica give me the derivative of P = 1/(1+Exp[-z]) as P(1-P) or in that form? Instead it gives me Exp[-z]/(1+Exp[-z])^2. Both are correct, but is there some way without by hand showing how the two solutions are equal that Mathematica could also provide me with that solution?
Thank you SOOO much!
Michelle
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