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  4. If y = xx , x > 0 , then (dy/dx) is

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

Limits and Derivatives

Solution:

$y =x^{x} \Rightarrow \log y =x \log x$
$ \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) $