A) \[{{x}^{x}}\log ex\]
B) \[{{x}^{x}}\left( 1+\frac{1}{x} \right)\]
C) \[(1+\log x)\]
D) \[{{x}^{x}}\log x\]
Correct Answer: A
Solution :
\[y={{x}^{x}}\] Taking \[\log \]on both sides, Þ \[\log y=x\log x\] Differentiating with respect to x, we get Þ \[\frac{1}{y}\frac{dy}{dx}=1+\log x\]; \[\therefore \frac{dy}{dx}={{x}^{x}}(1+\log x)={{x}^{x}}\log ex\].
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