If y = x^x + x^{frac{1}{x}},then frac{dy}{dx} | vedclass

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(A) Let $y = u + v$,where $u = x^x$ and $v = x^{\frac{1}{x}}$.Then $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$.For $u = x^x$,taking $\log$ on both

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$x^x(1+\log x) + x^{\frac{1}{x}} \frac{1}{x^2}(1-\log x)$

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(A) Let $y = u + v$,where $u = x^x$ and $v = x^{\frac{1}{x}}$.Then $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$.For $u = x^x$,taking $\log$ on both sides: $\log u = x \log x$.Differentiating with respect to $x$: $\frac{1}{u} \frac{du}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$.So,$\frac{du}{dx} = x^x(1 + \log x)$.For $v = x^{\frac{1}{x}}$,taking $\log$ on both sides: $\log v = \frac{1}{x} \log x$.Differentiating with respect to $x$: $\frac{1}{v} \frac{dv}{dx} = (-\frac{1}{x^2}) \log x + \frac{1}{x} (\frac{1}{x}) = \frac{1 - \log x}{x^2}$.So,$\frac{dv}{dx} = x^{\frac{1}{x}} \frac{1 - \log x}{x^2}$.Therefore,$\frac{dy}{dx} = x^x(1 + \log x) + x^{\frac{1}{x}} \frac{1 - \log x}{x^2}$.

If $y = x^x + x^{\frac{1}{x}}$,then $\frac{dy}{dx}$ is equal to

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