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JEE MATHEMATICS MCQ: DIFFERENTIATION

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Analyzing +4 | -1
For x>1\mathrm{x}>1, if (2x)2y=4e2x−2y(2 x)^{2 y}=4 e^{2 x-2 y}, then (1+log⁡e2x)2dydx\left(1+\log _{e} 2 x\right)^{2} \frac{d y}{d x} is equal to :
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Correct option: A
xlog⁡e2x−log⁡e2x\frac{x \log _{e} 2 x-\log _{e} 2}{x}
Option A (xlog⁡e2x−log⁡e2x\frac{x \log _{e} 2 x-\log _{e} 2}{x}) is correct because it is the result of differentiating the given equation and solving for the expression (1+log⁡e2x)2dydx(1+\log _{e} 2 x)^{2} \frac{d y}{d x}.

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