A Strange Derivative

Let $y = x^x$

$\begin{aligned} \Rightarrow \ln{y} & = x \ln{x} \\ \dfrac{1}{y} & = \ln{x} \dfrac {dx}{dy} + x\dfrac{1}{x} \dfrac{dx}{dy} \\ \dfrac{dx}{dy} & = \dfrac{1}{y(\ln{x}+1)} \\ \dfrac{d}{d\left( x^x\right)}x & = \dfrac{1}{x^x(\ln{x}+1)} \end{aligned}$

Challenge master ote -

$\frac d{d(x^x)} x = \frac d{d(x)} x \times \frac{1}{\frac{d(x^x)}{dx}}$

$y = x^x$

$logy = xlogx$

$\frac{dy}{dx} = x^x( logx + 1)$

$\frac d{d(x^x)} x = \frac{1}{ x^x( logx + 1)}$