For x>1,


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

$\left(1+\log _{e} 2 x\right)^{2} \frac{d y}{d x}$ is equal to :

  1. $\log _{e} 2 x$

  2. $\frac{x \log _{e} 2 x+\log _{e} 2}{x}$

  3. $x \log _{e} 2 x$

  4. $\frac{x \log _{e} 2 x-\log _{e} 2}{x}$

Correct Option: , 4


$(2 x)^{2 y}=4 e^{2 x-2 y}$

$2 \mathrm{y} \ell \mathrm{n} 2 \mathrm{x}=\ell \mathrm{n} 4+2 \mathrm{x}-2 \mathrm{y}$

$y=\frac{x+\ell n 2}{1+\ell n 2 x}$

$y^{\prime}=\frac{(1+\ell \operatorname{n} 2 x)-(x+\ell n 2) \frac{1}{x}}{(1+\ell n 2 x)^{2}}$

$y^{\prime}(1+\ell n 2 x)^{2}=\left[\frac{x \ell n 2 x-\ell n 2}{x}\right]$

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