How to compute the derivative of $\sqrt{x}^{\sqrt{x}}$? I know have the final answer and know I need to use the natural log but I'm confused about why that is.  

Could someone walk through it step by step?

Let $y= \sqrt{x}^{\sqrt x}$. Then $\ln(y) = \sqrt{x} \ln(\sqrt x) = \frac{1}{2} \sqrt x \ln(x)$. So,
$$
 \frac{d}{dx} \ln(y) = \frac{d}{dx} \left(\frac{1}{2}\sqrt x\ln(x)\right)
$$
$$
  \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left(\frac{\sqrt x}{x} + \frac{1}{2\sqrt x}\ln(x) \right)
$$
So, 
$$
  \frac{dy}{dx} = y \cdot \frac{1}{2} \left(\frac{\sqrt x}{x} + \frac{1}{2\sqrt x}\ln(x) \right) = \frac{1}{2}\sqrt{x}^{\sqrt x}\left(\frac{\sqrt x}{x} + \frac{1}{2\sqrt x}\ln(x) \right)
$$
