If y=(log x)^{1/x} + x^{log x},find frac{dy}{ | vedclass

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(C) Let $y = u + v$,where $u = (\log x)^{1/x}$ and $v = x^{\log x}$.Taking log on both sides for $u$: $\log u = \frac{1}{x} \log(\log x)$.Differentiat

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$\frac{1}{e^2} + 2$

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(C) Let $y = u + v$,where $u = (\log x)^{1/x}$ and $v = x^{\log x}$.Taking log on both sides for $u$: $\log u = \frac{1}{x} \log(\log x)$.Differentiating with respect to $x$: $\frac{1}{u} \frac{du}{dx} = \frac{1}{x} \cdot \frac{1}{\log x} \cdot \frac{1}{x} + \log(\log x) \cdot (-\frac{1}{x^2}) = \frac{1}{x^2 \log x} - \frac{\log(\log x)}{x^2}$.At $x=e$,$\log u = \frac{1}{e} \log(\log e) = \frac{1}{e} \log(1) = 0$,so $u = e^0 = 1$.Thus,$\frac{du}{dx} = u \left[ \frac{1}{e^2 \log e} - \frac{\log(\log e)}{e^2} \right] = 1 \left[ \frac{1}{e^2} - 0 \right] = \frac{1}{e^2}$.For $v = x^{\log x}$,taking log: $\log v = \log x \cdot \log x = (\log x)^2$.Differentiating: $\frac{1}{v} \frac{dv}{dx} = 2 \log x \cdot \frac{1}{x}$.At $x=e$,$\log v = (\log e)^2 = 1$,so $v = e^1 = e$.Thus,$\frac{dv}{dx} = v \left[ \frac{2 \log e}{e} \right] = e \left[ \frac{2}{e} \right] = 2$.Therefore,$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} = \frac{1}{e^2} + 2$.

If $y=(\log x)^{1/x} + x^{\log x}$,find $\frac{dy}{dx}$ at $x=e$.

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