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.
$\ln\ y=\sqrt{x} \ln\ x$
$\frac{dy}{dx}\cdot \frac{1}{y}=\sqrt{x}\ \frac{1}{x} +\ln\ x\ \frac{1}{2}x^{-\frac{1}{2}}$
$\sqrt{x}\cdot (\sqrt{x}\ \frac{1}{x} +\ln\ x \ \frac{1}{2}x^{-\frac{1}{2}})$
 A: That way could work though you made some mistakes, but an easier way shifts the square root to a fractional exponent.
$$\begin{align}
y&=\sqrt{x^x} \\[2ex]
 &= \left(x^x\right)^{1/2} \\[2ex]
 &= x^{x/2} \\[2ex]
\ln y&= \ln x^{x/2} \\[2ex]
 &= \frac x2\ln x \\[2ex]
\frac{dy}{dx}\frac 1y &=\frac 12\ln x+\frac x2\frac 1x \\[2ex]
 &= \frac 12\ln x+\frac 12 \\[2ex]
\frac{dy}{dx} &= y\left(\frac 12\ln x+\frac 12 \right) \\[2ex]
 &= \sqrt{x^x}\left(\frac 12\ln x+\frac 12 \right) \\[2ex]
 &= \frac 12\sqrt{x^x}\left(\ln x+1 \right)
\end{align}$$
A: Hint:
All functions of type $u^v$ are defined with:
$$u^v=\mathrm e^{v\ln u}.\enspace\text{Here:}\quad \sqrt{x^x}=\mathrm e^{\frac12 x\ln x}.$$
A: Square both sides:
$$y^{2}=x^{x}$$
Then upon differentiating
$$2y y' = (1+\ln x)x^{x}$$
From which
$$y'=\frac{1}{2y}(1+\ln x)x^{x} \qquad (x \neq 0)$$
Giving
$$y' = \frac{1}{2\sqrt{x^{x}}}(1+\ln x)x^{x} $$
Thus
$$y'=\frac{1}{2}(1+\ln x)\sqrt{x^{x}}$$
A: There are a couple of things that are not quite right, moreover taking the logarithm might not be the best choice (but it seems you are supposed to do this). 


*

*Note $\ln \sqrt{u}= \frac{1}{2} \ln u$, so your first line is incorrect.

*In the second you dropped an $x$ after the $\ln$. 

*In the third you took wrongly $\sqrt{x}$ for $y$.  
A: let $f(x)=\sqrt{x^x}$ then we obtain $\ln(f(x))=\frac{1}{2}\ln(x^x)=\frac{1}{2}x\ln(x)$ and after this we have $$\frac{1}{f}f'(x)=\frac{1}{2}\ln(x)+\frac{1}{2}$$
A: You made a few simple mistakes, most notably, you got the log wrong:
$$\ln y = \frac12 x \ln x$$
Differentiating then gives:
$$\frac{y'}{y}=\frac12\ln x+\frac12$$
And hence:
$$y'=y\left(\frac12\ln x +\frac12\right)=\frac12\sqrt{x^x}\left(\ln x+1\right)$$
