For x > 1, if (2x)2y = 4e2x$-$2y, then (1 + loge 2x)2 ${{dy} \over {dx}}$ is equal to :

Question

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

then (1 + log_e 2x)² ${{dy} \over {dx}}$ is equal to :

${{x\,{{\log }_e}2x - {{\log }_e}2} \over x}$

log_e 2x

x log_e 2x

${{x\,{{\log }_e}2x + {{\log }_e}2} \over x}$

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