For x 1,if (2x)^{2y} = 4e^{2x-2y},then (1 + | vedclass

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(D) Given the equation: $(2x)^{2y} = 4e^{2x-2y}$ Taking the natural logarithm on both sides: $2y \log_e(2x) = \log_e(4) + 2x - 2y$ $2y \log_e(2x) + 2y

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$\frac{x \log_e 2x - \log_e 2}{x}$

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(D) Given the equation: $(2x)^{2y} = 4e^{2x-2y}$ Taking the natural logarithm on both sides: $2y \log_e(2x) = \log_e(4) + 2x - 2y$ $2y \log_e(2x) + 2y = 2x + \log_e(4)$ $2y(1 + \log_e(2x)) = 2x + 2 \log_e(2)$ $y(1 + \log_e(2x)) = x + \log_e(2)$ $y = \frac{x + \log_e(2)}{1 + \log_e(2x)}$ Differentiating both sides with respect to $x$ using the quotient rule: $\frac{dy}{dx} = \frac{(1 + \log_e(2x)) \cdot \frac{d}{dx}(x + \log_e(2)) - (x + \log_e(2)) \cdot \frac{d}{dx}(1 + \log_e(2x))}{(1 + \log_e(2x))^2}$ $\frac{dy}{dx} = \frac{(1 + \log_e(2x)) \cdot 1 - (x + \log_e(2)) \cdot \frac{1}{2x} \cdot 2}{(1 + \log_e(2x))^2}$ $\frac{dy}{dx} = \frac{1 + \log_e(2x) - \frac{x + \log_e(2)}{x}}{(1 + \log_e(2x))^2}$ $(1 + \log_e(2x))^2 \frac{dy}{dx} = 1 + \log_e(2x) - 1 - \frac{\log_e(2)}{x}$ $(1 + \log_e(2x))^2 \frac{dy}{dx} = \log_e(2x) - \frac{\log_e(2)}{x} = \frac{x \log_e(2x) - \log_e(2)}{x}$

For $x > 1$,if $(2x)^{2y} = 4e^{2x-2y}$,then $(1 + \log_e 2x)^2 \frac{dy}{dx}$ is equal to

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