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

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(A) Given $(2x)^{2y} = 4e^{2x - 2y}$.Taking natural logarithm on both sides:$2y \ln(2x) = \ln(4) + 2x - 2y$$2y \ln(2x) + 2y = 2x + 2 \ln 2$$y(1 + \ln

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

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(A) Given $(2x)^{2y} = 4e^{2x - 2y}$.Taking natural logarithm on both sides:$2y \ln(2x) = \ln(4) + 2x - 2y$$2y \ln(2x) + 2y = 2x + 2 \ln 2$$y(1 + \ln 2x) = x + \ln 2$$y = \frac{x + \ln 2}{1 + \ln 2x} \quad \dots(1)$Now,differentiate both sides with respect to $x$:$\frac{dy}{dx} = \frac{(1 + \ln 2x)(1) - (x + \ln 2)(\frac{1}{x})}{(1 + \ln 2x)^2}$$\frac{dy}{dx} = \frac{1 + \ln 2x - 1 - \frac{\ln 2}{x}}{(1 + \ln 2x)^2}$$\frac{dy}{dx} = \frac{\ln 2x - \frac{\ln 2}{x}}{(1 + \ln 2x)^2}$Multiply both sides by $(1 + \ln 2x)^2$:$(1 + \ln 2x)^2 \frac{dy}{dx} = \ln 2x - \frac{\ln 2}{x}$$(1 + \ln 2x)^2 \frac{dy}{dx} = \frac{x \ln 2x - \ln 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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