For x > 1, if (2x)2y = 4e(2x-2y), then \({\left( {1 + lo{g_e}2x} \right)^2}\frac{{dy}}{{dx}}\) is equal to

\(\frac{{(xlo{g_e}{\rm{\;}}2x + lo{g_e}{\rm{\;}}2}}{x}\)

\(\frac{{(xlo{g_e}{\rm{}}2x - lo{g_e}{\rm{\;}}2}}{x}\)

For x > 1, if (2x)2y = 4e(2x-2y), then \({\left( {1 + lo{g_e}2x} \ri

1.	For x > 1, if (2x)2y = 4e(2x-2y), then \({\left( {1 + lo{g_e}2x} \right)^2}\frac{{dy}}{{dx}}\) is equal to
A.	\(\frac{{(xlo{g_e}{\rm{\;}}2x + lo{g_e}{\rm{\;}}2}}{x}\)
B.	\(\frac{{(xlo{g_e}{\rm{}}2x - lo{g_e}{\rm{\;}}2}}{x}\)
C.	x loge 2x
D.	loge 2x
Answer» C. x loge 2x

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