If y=x^{log x}+(log x)^x, x1 then left(frac{ | vedclass

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(D) Let $y = u + v$, where $u = x^{\log x}$ and $v = (\log x)^x$. Taking log on both sides for $u$: $\log u = (\log x)(\log x) = (\log x)^2$. Differen

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$3$

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(D) Let $y = u + v$, where $u = x^{\log x}$ and $v = (\log x)^x$. Taking log on both sides for $u$: $\log u = (\log x)(\log x) = (\log x)^2$. Differentiating with respect to $x$: $\frac{1}{u} \frac{du}{dx} = 2(\log x) \cdot \frac{1}{x} \implies \frac{du}{dx} = x^{\log x} \cdot \frac{2 \log x}{x}$. At $x=e$: $\frac{du}{dx} = e^{\log e} \cdot \frac{2 \log e}{e} = e^1 \cdot \frac{2}{e} = 2$. Now for $v = (\log x)^x$: $\log v = x \log(\log x)$. Differentiating with respect to $x$: $\frac{1}{v} \frac{dv}{dx} = 1 \cdot \log(\log x) + x \cdot \frac{1}{\log x} \cdot \frac{1}{x} = \log(\log x) + \frac{1}{\log x}$. $\frac{dv}{dx} = (\log x)^x \left[ \log(\log x) + \frac{1}{\log x} \right]$. At $x=e$: $\frac{dv}{dx} = (\log e)^e \left[ \log(\log e) + \frac{1}{\log e} \right] = 1^e [ \log(1) + 1 ] = 1 \cdot [0 + 1] = 1$. Thus, $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} = 2 + 1 = 3$.

If $y=x^{\log x}+(\log x)^x, x>1$ then $\left(\frac{d y}{d x}\right)_{x=e}=$

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