QJEEM19Mae691: JEE Main 2019 Mathematics - Class 12 JEEM19Mae691 Solution: For x > 1, if (2x) 2y = 4e 2x − 2y , then (1 + log e 2x) 2 d y…

Question

Verified Solution • 2025–26 Session moderate

For x > 1, if (2x)^2y = 4e^{2x $-$ 2y},

then (1 + log_e 2x)²

\frac{d y}{d x}

is equal to :

Accepted Answer

(2x)^2y = 4e^2x-2y

2y

ℓ

n2x =

ℓ

n4 + 2x

-

2y

y =

\frac{x + ℓ n 2}{1 + ℓ n 2 x}

y ' =

\frac{(1 + ℓ n 2 x) - (x + ℓ n 2) \frac{1}{x}}{{(1 + ℓ n 2 x)}^{2}}

y '

{(1 + ℓ n 2 x)}^{2} = [\frac{x ℓ n 2 x - ℓ n 2}{x}]

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