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(2x)^(2y) = 4e^(2x - 2y): find (1 + log 2x)^2 dy/dx

The same question, asked 2 times: 10 May 2024 S1, 12 May 2024 S2. Every wording below, oldest first.

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  1. Question 1

    10 May 2024 S1

    124
    For x>1x>1, (2x)2y=4e2x−2y(2x)^{2y}=4e^{2x-2y}. Value of (1+log⁡2x)2dydx(1+\log2x)^2\frac{dy}{dx} is

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  2. Question 2

    12 May 2024 S2

    128
    For x>1x > 1, if (2x)2y=4e2x−2y(2x)^{2y} = 4e^{2x-2y}, then (1+log⁡e2x)2dydx\left(1+\log_e 2x\right)^2 \frac{dy}{dx} is equal to

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