What is the derivative of a super square root?

Question

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. Can somebody give me a explanation for that? Thanks.

Answer 1

$$y^y=x$$

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 $x$.