If you mean: y =(lnx)3
then:
dy/dx = [3(lnx)2]/x
ddy/dx = [(6lnx / x) - 3(lnx)2] / x2
If you mean: y = ln(x3)
Then:
dy/dx = 3x2/x3 = 3/x = 3x-1
ddy/dx = -3x-2 = -3/x2
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Oh, dude, the third derivative of ln(x) is -2/(x^3). But like, who really needs to know that, right? I mean, unless you're planning on impressing your calculus teacher or something. Just remember, math is like a puzzle, except no one actually wants to put it together.
First one: f'(x) = 2x*lnx + x^2*(1/x) = 2x*lnx + x = x*(2*lnx + 1) Second derivate: f"(x) = D [x*(2*lnx + 1)] = 1*(2*lnx + 1) + x*(2/x) = 2*lnx+1+2 = 2*lnx + 3 So, there is a minimum in this graph on point (1/e^(1/2)), -1/(2e)) = MIN(x, y) Van Sanden David Belgium
(xlnx)' = lnx + 1
There are 2 interpretations of your question: First: e^[lnx + lny] =e^[ln(xy)] =xy Second: lny + e^(lnx) =lny + x