A) \[\frac{x\,{{\log }_{e}}2x-{{\log }_{e}}2}{x}\]
B) \[{{\log }_{e}}2x\]
C) \[\frac{x\,{{\log }_{e}}2x+{{\log }_{e}}2}{x}\]
D) \[x\,{{\log }_{e}}2x\]
Correct Answer: A
Solution :
\[{{(2x)}^{2y}}=4{{e}^{2x-2y}}\] \[\Rightarrow \]\[2y{{\log }_{e}}2x={{\log }_{e}}4+2x-2y\] \[\Rightarrow \]\[2y{{\log }_{e}}2x=2{{\log }_{e}}2+2x-2y\] \[\Rightarrow \]\[y{{\log }_{e}}2x={{\log }_{e}}2+x-y\Rightarrow y=\frac{x+{{\log }_{e}}2}{1+{{\log }_{e}}2x}\] \[\therefore \]\[\frac{dy}{dx}=\frac{1+{{\log }_{e}}2x-(x+lo{{g}_{e}}2)\frac{1}{x}}{{{(1+lo{{g}_{e}}2x)}^{2}}}\] \[\Rightarrow \]\[\frac{dy}{dx}={{(1+lo{{g}_{e}}2x)}^{2}}=\frac{x{{\log }_{e}}2x-{{\log }_{e}}2}{x}\]
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