1)

$\displaystyle

(x + \frac {1}{x})' = \frac {1}{2}x^{-0.5} - \frac {x^{0.5}}{x}= \frac {1}{2\sqrt{x}} - \frac {\sqrt{x}}{x} = \frac{\sqrt{x}}{2x}

$

The solution is supposed to be $\displaystyle \frac{\sqrt{x}(x-1)}{2x^2}$

2)

$\displaystyle

(ln \sqrt{\frac{a+x}{a-x}})' = \frac{a-x}{2(a+x)}

$

The solution is supposed to be $\displaystyle \frac{a}{a^2-x^2}$