<s>[INST] Differentiate the Function: $y=\sqrt{x^x}$ $y=\sqrt{x^x}$
How do I convert this into a form that is workable and what indicates that I should do so? 
Anyway, I tried this method of logging both sides of the equation but I don't know if I am right.
$\ln\ y=\sqrt{x} \ln\ x$
$\frac{dy}{dx}\cdot \frac{1}{y}=\sqrt{x}\ \frac{1}{x} +\ln\ x\ \frac{1}{2}x^{-\frac{1}{2}}$
$\sqrt{x}\cdot (\sqrt{x}\ \frac{1}{x} +\ln\ x \ \frac{1}{2}x^{-\frac{1}{2}})$
 [/INST] That way could work though you made some mistakes, but an easier way shifts the square root to a fractional exponent.
$$\begin{align}
y&=\sqrt{x^x} \\[2ex]
 &= \left(x^x\right)^{1/2} \\[2ex]
 &= x^{x/2} \\[2ex]
\ln y&= \ln x^{x/2} \\[2ex]
 &= \frac x2\ln x \\[2ex]
\frac{dy}{dx}\frac 1y &=\frac 12\ln x+\frac x2\frac 1x \\[2ex]
 &= \frac 12\ln x+\frac 12 \\[2ex]
\frac{dy}{dx} &= y\left(\frac 12\ln x+\frac 12 \right) \\[2ex]
 &= \sqrt{x^x}\left(\frac 12\ln x+\frac 12 \right) \\[2ex]
 &= \frac 12\sqrt{x^x}\left(\ln x+1 \right)
\end{align}$$
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