If x cdot log _{e}(log _{e} x)-x^2+y^2=4 and | vedclass

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(B) Given the equation: $x \cdot \log_{e}(\log_{e} x) - x^2 + y^2 = 4$. First,find the value of $y$ at $x=e$: $e \cdot \log_{e}(\log_{e} e) - e^2 + y^

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$\frac{2e-1}{2\sqrt{4+e^2}}$

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(B) Given the equation: $x \cdot \log_{e}(\log_{e} x) - x^2 + y^2 = 4$. First,find the value of $y$ at $x=e$: $e \cdot \log_{e}(\log_{e} e) - e^2 + y^2 = 4$. Since $\log_{e} e = 1$ and $\log_{e} 1 = 0$,we have $e \cdot 0 - e^2 + y^2 = 4$,which implies $y^2 = 4 + e^2$. Since $y > 0$,$y = \sqrt{4 + e^2}$. Now,differentiate the given equation with respect to $x$: $\frac{d}{dx}[x \cdot \log_{e}(\log_{e} x)] - \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = 0$. Using the product rule on the first term: $1 \cdot \log_{e}(\log_{e} x) + x \cdot \frac{1}{\log_{e} x} \cdot \frac{1}{x} - 2x + 2y \frac{dy}{dx} = 0$. $\log_{e}(\log_{e} x) + \frac{1}{\log_{e} x} - 2x + 2y \frac{dy}{dx} = 0$. Substitute $x=e$: $\log_{e}(\log_{e} e) + \frac{1}{\log_{e} e} - 2e + 2y \frac{dy}{dx} = 0$. $\log_{e}(1) + \frac{1}{1} - 2e + 2y \frac{dy}{dx} = 0$. $0 + 1 - 2e + 2y \frac{dy}{dx} = 0$. $2y \frac{dy}{dx} = 2e - 1$. $\frac{dy}{dx} = \frac{2e - 1}{2y}$. Substituting $y = \sqrt{4 + e^2}$,we get $\frac{dy}{dx} = \frac{2e - 1}{2\sqrt{4 + e^2}}$.

If $x \cdot \log _{e}(\log _{e} x)-x^2+y^2=4$ and $y>0$,then $\frac{dy}{dx}$ at $x=e$ is

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