How to differentiate f(x)=e^(xlnx)?
I get f'(x)=e^(xlnx) (1+lnx) while at wolfram is x^x(ln x+x)...sorry for my miscalculation / typo error...
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Your answer is correct, since e^(x lnx) = (e^(lnx))^x = x^x
I get f'(x)=e^(xlnx) (1+lnx) while at wolfram is x^x(ln x+x)...sorry for my miscalculation / typo error...
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