If x ln(ln x) - x^2 + y^2 = 4 where y 0,the | vedclass

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(B) Given the equation: $x \ln(\ln x) - x^2 + y^2 = 4$.Differentiating both sides with respect to $x$:$\frac{d}{dx}[x \ln(\ln x)] - \frac{d}{dx}(x^2)

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

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(B) Given the equation: $x \ln(\ln x) - x^2 + y^2 = 4$.Differentiating both sides with respect to $x$:$\frac{d}{dx}[x \ln(\ln x)] - \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(4)$Using the product rule on the first term: $1 \cdot \ln(\ln x) + x \cdot \frac{1}{\ln x} \cdot \frac{1}{x} - 2x + 2y \frac{dy}{dx} = 0$$\ln(\ln x) + \frac{1}{\ln x} - 2x + 2y \frac{dy}{dx} = 0$At $x = e$,$\ln(\ln e) = \ln(1) = 0$ and $\ln e = 1$:$0 + \frac{1}{1} - 2e + 2y \frac{dy}{dx} = 0$$1 - 2e + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{2e - 1}{2y}$Now,find $y$ at $x = e$ from the original equation:$e \ln(\ln e) - e^2 + y^2 = 4$$e(0) - e^2 + y^2 = 4 \implies y^2 = 4 + e^2 \implies y = \sqrt{4 + e^2}$ (since $y > 0$)Substituting $y$ into the expression for $\frac{dy}{dx}$:$\frac{dy}{dx} = \frac{2e - 1}{2\sqrt{4 + e^2}}$

If $x \ln(\ln x) - x^2 + y^2 = 4$ where $y > 0$,then $\frac{dy}{dx}$ at $x = e$ is equal to

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