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) \[\frac{{{e}^{x}}[1+(x-2)\log x]}{{{x}^{3}}}\]
Correct Answer: D
Solution :
\[\frac{dy}{dx}=-2{{x}^{-3}}{{e}^{x}}\log x+{{x}^{-2}}\left( {{e}^{x}}\log x+\frac{{{e}^{x}}}{x} \right)\] \[={{e}^{x}}\left[ \frac{1+(x-2)\log x}{{{x}^{3}}} \right]\] Aliter: Taking \[\log \], \[\log y=x+\log \log x-2\log x\] Þ \[\frac{1}{y}\frac{dy}{dx}=1+\frac{1}{x\log x}-\frac{2}{x}\] Þ \[\frac{dy}{dx}=\frac{{{e}^{x}}\log x}{{{x}^{2}}}\], \[\left[ \frac{x\log x+1-2\log x}{x\log x} \right]\] \[=\frac{{{e}^{x}}[(x-2)\log x+1]}{{{x}^{3}}}\].
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