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Math Help - logarithms, confusing

  1. #1
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    logarithms, confusing

    Given that y = X^x , x>0, y>0, by taking logs, show that

    dy/dx = (X^x)(1+lnx)
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  2. #2
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    Quote Originally Posted by Stylis10 View Post
    Given that y = X^x , x>0, y>0, by taking logs, show that

    dy/dx = (X^x)(1+lnx)
    Rewriting using logs
    lny=xlnx

    Taking the derivative
    \frac{1}{y}\frac{dy}{dx}=lnx+x\frac{1}{x}

    solving for y' gives

    \frac{dy}{dx}=y(lnx+1)

    sub in what y is equal to and vola!

    \frac{dy}{dx}=x^x(lnx+1)
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  3. #3
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    Quote Originally Posted by Stylis10 View Post
    Given that y = X^x , x>0, y>0, by taking logs, show that

    dy/dx = (X^x)(1+lnx)
    y = x^x ............. Calculate the logarithms of both sides:

    \ln(y) = \ln(x^x) = x \cdot \ln(x) ............. Differentiate. Use the chainrule at the LHS and the product rule at the RHS:

    \frac1y \cdot \frac{dy}{dx}  = x \cdot \frac1x + \ln(x) \cdot 1 ............. Simplify:

     \frac{dy}{dx}  = y \cdot (1 + \ln(x)) Substitute y by the term of the function:

     \frac{dy}{dx}  = x^x \cdot (1 + \ln(x))
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