Question:

If $y = x^x , x > 0 $ , then $\frac{dy}{dx}$ is

Updated On: Jul 6, 2022
  • log x
  • 2 + log x
  • $x^x \log x $
  • $x^x ( 1 + \log x )$
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The Correct Option is D

Solution and Explanation

$y =x^{x} \Rightarrow \log y =x \log x$ $ \Rightarrow \frac{1}{y} \frac{dy}{dx} = x. \frac{1}{x} + \log x.1$ $ \Rightarrow \frac{dy}{dx} = y\left(1+\log x\right) =x^{x} \left(1+\log x\right) $
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