If x^x y^y=e^e,then left(frac{d^2 y}{d x^2}ri | vedclass

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(A) Given $x^x y^y=e^e$. Taking natural logarithm on both sides:$\ln(x^x y^y) = \ln(e^e)$$x \ln x + y \ln y = e$Differentiating with respect to $x$:$\

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$\frac{1}{e}\left(\frac{d y}{d x}\right)_{(e, e)}$

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(A) Given $x^x y^y=e^e$. Taking natural logarithm on both sides:$\ln(x^x y^y) = \ln(e^e)$$x \ln x + y \ln y = e$Differentiating with respect to $x$:$\frac{d}{dx}(x \ln x) + \frac{d}{dx}(y \ln y) = \frac{d}{dx}(e)$$(1 + \ln x) + (1 + \ln y) \frac{dy}{dx} = 0$At point $(e, e)$,$\ln e = 1$:$(1 + 1) + (1 + 1) \left(\frac{dy}{dx}\right)_{(e, e)} = 0$$2 + 2 \left(\frac{dy}{dx}\right)_{(e, e)} = 0 \Rightarrow \left(\frac{dy}{dx}\right)_{(e, e)} = -1$Differentiating $(1 + \ln x) + (1 + \ln y) y' = 0$ again with respect to $x$:$\frac{1}{x} + \frac{1}{y} y' \cdot y' + (1 + \ln y) y'' = 0$$\frac{1}{x} + \frac{(y')^2}{y} + (1 + \ln y) y'' = 0$At $(e, e)$ where $y' = -1$:$\frac{1}{e} + \frac{(-1)^2}{e} + (1 + \ln e) y'' = 0$$\frac{1}{e} + \frac{1}{e} + 2 y'' = 0$$\frac{2}{e} + 2 y'' = 0 \Rightarrow y'' = -\frac{1}{e}$Since $\left(\frac{dy}{dx}\right)_{(e, e)} = -1$,we can write $y'' = \frac{-1}{e} = \frac{1}{e} \left(\frac{dy}{dx}\right)_{(e, e)}$.

If $x^x y^y=e^e$,then $\left(\frac{d^2 y}{d x^2}\right)_{(e, e)}=$

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