解答

$${y=\log(\sqrt{x}+1)}$$

$$\begin{eqnarray*} y'&=&\frac{(\sqrt{x}+1)'}{\sqrt{x}+1}\\ &=& \frac{\frac{1}{2\sqrt{x}}}{\sqrt{x}+1}\\ &=& \frac{1}{2\sqrt{x}(\sqrt{x}+1)} \end{eqnarray*}$$