Uppgift 2

Derivatans definition ger för $x>0$ att
$$\begin{aligned}f'(x)&=\lim_{h\to0}\frac{f(x+h)-f(x)}h=\lim_{h\to0}\frac1h\left(\frac1{\sqrt{x+h}+1}-\frac1{\sqrt x+1}\right)\\&=\lim_{h\to0}\frac{\sqrt x-\sqrt{x+h}}{h(\sqrt{x+h}+1)(\sqrt x+1)}=[\text{förlänger med konjugat}]\\&=\lim_{h\to0}\frac{-1}{(\sqrt{x+h}+1)(\sqrt x+1)(\sqrt x+\sqrt{x+h})}\\&=\frac{-1}{(\sqrt x+1)(\sqrt x+1)(\sqrt x+\sqrt x)}=\frac{-1}{2\sqrt x(\sqrt x+1)^2}.\end{aligned}$$