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?
 A: 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, 
$$
  \frac{dy}{dx} = y \cdot \frac{1}{2} \left(\frac{\sqrt x}{x} + \frac{1}{2\sqrt x}\ln(x) \right) = \frac{1}{2}\sqrt{x}^{\sqrt x}\left(\frac{\sqrt x}{x} + \frac{1}{2\sqrt x}\ln(x) \right)
$$
A: The idea is to turn the problem into something to which one can apply the standard rules ( product and chain rules, for instance) to functions the derivative of which one already knows. Now, in your case you have a function raised to another function - which is complicated: there is no 'exponentiation rule' for derivatives, as such (i.e., both the base and exponent are not constants - you know how to take the derivative of $x^n$  and $ e^x $ - but not  of $x^x$, at least without some manipulation); but 
$$ \ln ( a ^ b ) = b \ln a $$
i.e., if you introduce a logarithm into the equation, the exponentiation becomes multiplication - so we can hope to use the product rule.
Now, you do not have a $\ln$ in the question - if you introduce one, you must add at the same time, its inverse, an exponentiation:
$$ a^b = e^{\ln a^b } = e ^ { b \ln a }$$
While this might look like we are complicating things, we are not: we know how to take the derivative of "e to the something" and the derivative of the "ln of something" - and the weird exponentiation has become a product, for which there is a differentiation rule...
So - write 
$$ (\sqrt x )^{\sqrt x} = e ^ { \ln ( { \sqrt x } ^ { \sqrt x } )} 
                        = e^ { \sqrt x \ln ( \sqrt x) } $$ 
- the derivative of the function on the right will be the derivative of the function on the left (i.e., of the one you want).
Now, the derivative of 
$$ e^{whatever}$$
is, by the chain rule,
$$ e ^ { whatever } ( derivative\ of\ whatever ). $$
Here 'whatever' is  $$ \sqrt x \ln ( \sqrt x )$$
Are you OK, at this point - or do you need more help?
*Edit - For completeness sake, although at the risk of making this long answer too long... Other answers here (contrary, perhaps, to appearances) are doing the same thing as this one; the point is to use a logarithm to turn an exponentiation into a multiplication: if 
$$ y = {f(x)}^{g(x)},$$
then
$$ \ln y  = \ln \left({f(x)}^{g(x)}\right) = g(x) \ln f ( x ) ,$$
i.e., the original exponentiation on the right of the equal side becomes a product - so the product rule now applies. 
In your case, $g(x) = f(x) = \sqrt x$... Be that as it may, $g(x) \ln ( f(x) )$ is what was called 'whatever' above.
Taking the derivative on both sides of the equation - and remembering the chain rule -  one gets (writing it to make it look like the first version in this answer)
$$   {y'\over y} = derivative\ of\ whatever.$$
Cross-multiplying with $y$, one obtains
$$ y' = y \ \cdot \  (derivative\ of\ whatever),$$
as was the case in the 'logarithmic derivative' answers - but also in this answer! To be explicit, with your $y$, 
$$ y =  {\sqrt x}^{\sqrt x} = e^{\sqrt x \ln \sqrt x }=e^{whatever},$$
so the methods are the same.
To summarize: (either version of) the method is useful if there is an exponentiation with both base and exponent not constant.
A: let $y=\sqrt{x}^\sqrt{x}$, then $\ln{y}=\sqrt{x}\ln{\sqrt{x}}=\frac{1}{2}x^{\frac{1}{2}}\ln{x}$. Then take $d(.)$ on both sides, we have:
$$
\frac{dy}{y}=\frac{1}{2}(\frac{1}{2}x^{-\frac{1}{2}}\ln{x}+x^{\frac{1}{2}}\frac{1}{x})dx
$$
$$
\frac{dy}{dx}=\frac{y}{2}x^{-\frac{1}{2}}(1+\frac{1}{2}\ln{x})=\frac{x^{\sqrt{x}-1}}{2}(1+\frac{1}{2}\ln{x})
$$
A: Notice that $\sqrt{x}^{\sqrt{x}}=x^{\frac{1}{2}\sqrt{x}}$. Applying the chain rule yields the derivative: $$\left(\frac{1}{2}\sqrt{x}\right)\cdot x^{\frac{1}{2}\sqrt{x}-1}+(\ln x)\cdot \frac{1}{2}\cdot \frac{1}{2\sqrt{x}}\cdot  x^{\frac{1}{2}\sqrt{x}}.$$
