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

dy/dx = (X^x)(1+lnx)

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- Feb 24th 2008, 07:18 AMStylis10logarithms, confusing
Given that y = X^x , x>0, y>0, by taking logs, show that

dy/dx = (X^x)(1+lnx) - Feb 24th 2008, 07:29 AMTheEmptySet
- Feb 24th 2008, 07:31 AMearboth
$\displaystyle y = x^x$ ............. Calculate the logarithms of both sides:

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

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

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

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