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OCW Scholar - Single Variable Calculus OpenStudy (anonymous):

derivate x^2x OpenStudy (anonymous):

$y=x ^{2x}$ $\ln y=\ln x ^{2x}$ $\ln y=2x.\ln x$ $(\ln y)\prime=(2x.\ln x)\prime$ $y \prime/y=2.\ln x+1/x.2x$ $y \prime=(2.\ln x+2).y$ $y \prime=2.y.(\ln x+1)$ $y \prime=2.x ^{2x}.(\ln x+1)$ OpenStudy (tad1):

Let u = 2x. then du /dx = 2 then dy/dx = dy/du/du/dx = u *x^(u-1) du/dx = 2(2x)x^(2x-1)= 4x^2x) OpenStudy (anonymous):

1st answer might be correct....2nd answer i dont think so because it was in my test worth 6marks....so i cant get 6 marks for such simple stuff there is a trick to this queestion OpenStudy (tad1):

You're right. There is a trick. f = x^2x we know that a^x = e^(e^xlna) let u = e^2xlnx; du/dx = (e^2x)/x + (2e^2x)ln x so f' =dy/du)(du/dx)= [e^(xlna)] (ln a) = a^x lna} du/dx for our problem it would be (x^2x ln 2x)[(e^2x)/x + 2e^2x)ln x) I think. OpenStudy (anonymous):

tad is wrong the easiest way to do so is to do this $e^(2xlnx)=e^(lny)$ then differinitiate the left side only e^2xlnx(2lnx+2) then you get x^2x(2ln2x+2)=dy/dx because the e^ln cancels you just get y prime hope this helps

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