Re: Differentiation error

Jordan Watts wrote:
> If you derive e^x, instead of just getting e^x back, you get e^x*ln(e). I can
> see this is the same, but why does it do it?

In short it is because Graph differentiates the power function in a 
general way. This results in f'(x)=x*e^(x-1)*0+e^x*ln(e)*1. Afterwards 
Graph optimizes away multiplications by 1 and 0. I will add to the 
optimizer so it also removed the ln(e).

Best regards
Ivan Johansen

-------------------------------------------------------------------------
This SF.net email is sponsored by: Microsoft
Defy all challenges. Microsoft(R) Visual Studio 2008.
http://clk.atdmt.com/MRT/go/vse0120000070mrt/direct/01/