Implicit derivatives and logarithmic derivatives: $\left(x^{\sqrt{x}}\right)'=?$ How would I find the derivative with respect to $x$ of $$y = \left(x^{\sqrt{x}}\right)'.$$ I can find the correct answer using the method of logarithmic differentiation that my book  mysteriously suggest to use for some reason, BUT how can I find the derivative without using that method?
Furthermore how would I know when I should use logarithmic differentiation? 
And why can I not get the correct answer using the Chain rule only? 

Alternatively, using a logarithmic derivative as was suggested to you, from $$y=x^{\sqrt{x}}=e^{\sqrt{x}\ln x}$$
we have $$
(\ln y)'=\frac{y'}y=(\sqrt{x}\ln x)'=\frac{\ln x}{2\sqrt{x}}+\frac{\sqrt{x}}x
$$ that is
$$
\frac{(x^{\sqrt{x}})'}{x^{\sqrt{x}}}=\frac{\ln x}{2\sqrt{x}}+\frac1{\sqrt{x}}
$$ giving

$$
\left(x^{\sqrt{x}}\right)'=\left(\frac{\ln x}2+1\right)x^{\sqrt{x}-\frac12},\qquad x>0.
$$

