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y = xx
Haha yeah, I got it immediately after I posted.y = xx
.: lny = ln(xx) = xlnx
.: d/dx both sides: (1/y)dy/dx = x/x + lnx = 1 + lnx
,: dy/dx = y(1+lnx) = (1+lnx)xx
Good one, gurmies.y = x^x = e^ln(x^x) = e^(xlnx)
dy/dx = (1+lnx)*e^(xlnx)
= x^x*(1+lnx)
This negates the need to use implicit differentation.