# Derivative of super square root [closed]

What is the derivative of $y=^{1/2}x$? I tried finding the derivative of $x^{x}$ and then finding the inverse of that, but that didn't work.

## closed as off-topic by user21820, Jyrki Lahtonen, Brian Borchers, Jens, Ethan BolkerAug 8 '18 at 22:11

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$$y^y=x$$

for $x,y>0$

$$y\ln y=\ln x$$

Differentiating both the sides :

$$1 \cdot \ln y \cdot y'+y \cdot {\frac 1y} \cdot y' = \frac 1x$$

$$y'=\frac{1}{x(\ln y +1)}$$

$$\frac {dy}{dx}= \frac {1}{x(\ln y +1)}$$

Where $y$ is super square-root of $x$.