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Math Help - Derivative of e^(xlnx)

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    Derivative of e^(xlnx)

    For the question e^(xlnx), the derivative is given as x^x(1+lnx). I understand where the (1+lnx) comes from, that is the derivative of xlnx, but I'm not sure why it's being multiplied by x^x. I thought that d/dx(e^u) = u' e^u

    Second, for the question y=lnx at the point where x=e/2, find the equation of the tangent at that point. Isn't the derivative and slope here 1/2e? The equation given as the answer is y = (2/e)x
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by theowne View Post
    For the question e^(xlnx), the derivative is given as x^x(1+lnx). I understand where the (1+lnx) comes from, that is the derivative of xlnx, but I'm not sure why it's being multiplied by x^x. I thought that d/dx(e^u) = u' e^u
    you are correct in your last formula.

    so doing it directly, you would obtain, (1 + \ln x)e^{x \ln x}

    but you could also realize that e^{x \ln x} = e^x e^{\ln x} = xe^x

    and find the derivative of that by the product rule

    Second, for the question y=lnx at the point where x=e/2, find the equation of the tangent at that point. Isn't the derivative and slope here 1/2e? The equation given as the answer is y = (2/e)x
    the slope is not what you say

    the derivative is 1/x, if you plug in x = e/2, you get the slope being 2/e
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    For the first one, note that e^{xlnx} = e^{{lnx}^x}.

    Recall that e^{lnx} = x and you get x^x.
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    Quote Originally Posted by Jhevon View Post
    (1 + \ln x)e^{x \ln x}

    but you could also realize that e^{x \ln x} = e^x e^{\ln x} = xe^x
    No: e^x e^{\ln x} = e^{x + lnx}, not x^{xlnx}
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathnasium View Post
    No: e^x e^{\ln x} = e^{x + lnx}, not x^{xlnx}
    correct. sorry about that. was a little distracted.

    i still think the formula the asker posted first is the easier way though
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    e^{x\ln (x)}  = e^{\ln \left( {x^x } \right)}  = x^x \,\& \,y = x^x  \Rightarrow \quad y' = x^x \left( {1 + \ln (x)} \right)<br />
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Plato View Post
    e^{x\ln (x)}  = e^{\ln \left( {x^x } \right)}  = x^x \,\& \,y = x^x  \Rightarrow \quad y' = x^x \left( {1 + \ln (x)} \right)<br />
    just to add. Plato used logarithmic differentiation here
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