For x1,if (2 x)^{2 y}=4 e^{2 x-2 y},then (1+ | vedclass

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

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

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(B) Given equation: $(2 x)^{2 y}=4 e^{2 x-2 y}$.Taking natural logarithm on both sides:$2 y \log(2 x) = \log(4) + \log(e^{2 x-2 y})$$2 y \log(2 x) = 2 \log 2 + 2 x - 2 y$Dividing by $2$:$y \log(2 x) = \log 2 + x - y$$y(1 + \log(2 x)) = x + \log 2$$y = \frac{x + \log 2}{1 + \log(2 x)}$Differentiating with respect to $x$ using the quotient rule:$\frac{d y}{d x} = \frac{(1 + \log(2 x)) \cdot \frac{d}{d x}(x + \log 2) - (x + \log 2) \cdot \frac{d}{d x}(1 + \log(2 x))}{(1 + \log(2 x))^2}$$\frac{d y}{d x} = \frac{(1 + \log(2 x)) \cdot 1 - (x + \log 2) \cdot \frac{1}{2 x} \cdot 2}{(1 + \log(2 x))^2}$$\frac{d y}{d x} = \frac{1 + \log(2 x) - \frac{x + \log 2}{x}}{(1 + \log(2 x))^2}$Multiplying by $(1 + \log(2 x))^2$:$(1 + \log(2 x))^2 \frac{d y}{d x} = 1 + \log(2 x) - 1 - \frac{\log 2}{x}$$= \log(2 x) - \frac{\log 2}{x} = \frac{x \log(2 x) - \log 2}{x}$.

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

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