If \[y=\frac{{{e}^{x}}\log x}{{{x}^{2}}}\], then \[\frac{dy}{dx}=\] [AI CBSE 1982]

\[\frac{{{e}^{x}}[1+(x+2)\log x]}{{{x}^{3}}}\]

\[\frac{{{e}^{x}}[1-(x-2)\log x]}{{{x}^{4}}}\]

\[\frac{{{e}^{x}}[1-(x-2)\log x]}{{{x}^{3}}}\]

\[\frac{{{e}^{x}}[1+(x-2)\log x]}{{{x}^{3}}}\]

If \[y=\frac{{{e}^{x}}\log x}{{{x}^{2}}}\], then \[\frac{dy}{dx}=\]�

1.	If \[y=\frac{{{e}^{x}}\log x}{{{x}^{2}}}\], then \[\frac{dy}{dx}=\] [AI CBSE 1982]
A.	\[\frac{{{e}^{x}}[1+(x+2)\log x]}{{{x}^{3}}}\]
B.	\[\frac{{{e}^{x}}[1-(x-2)\log x]}{{{x}^{4}}}\]
C.	\[\frac{{{e}^{x}}[1-(x-2)\log x]}{{{x}^{3}}}\]
D.	\[\frac{{{e}^{x}}[1+(x-2)\log x]}{{{x}^{3}}}\]
Answer» E.