2020-M2-02

$\begin{array}{cl}
& f'(2) \\
= & \dlim_{h\to 0} \dfrac{f(2+h)-f(2)}{h} \\
= & \dlim_{h\to 0} \dfrac{1}{h} \left( \dfrac{2+h}{\sqrt{2+2+h}}-\dfrac{2}{\sqrt{2+2}}\right) \\
= & \dlim_{h\to 0} \dfrac{1}{h} \left( \dfrac{2+h}{\sqrt{4+h}}-1 \right) \\
= & \dlim_{h\to 0} \dfrac{2+h-\sqrt{4+h}}{h\sqrt{4+h}} \\
= & \dlim_{h\to 0} \dfrac{(2+h)-\sqrt{4+h}}{h\sqrt{4+h}} \times \dfrac{(2+h)+\sqrt{4+h}}{(2+h)+\sqrt{4+h}} \\
= & \dlim_{h\to 0} \dfrac{(2+h)^2-(\sqrt{4+h})^2}{h\sqrt{4+h}(2+h+\sqrt{4+h})} \\
= & \dlim_{h\to 0} \dfrac{4+4h+h^2-4-h}{h\sqrt{4+h}(2+h+\sqrt{4+h})} \\
= & \dlim_{h\to 0} \dfrac{h^2+3h}{h\sqrt{4+h}(2+h+\sqrt{4+h})} \\
= & \dlim_{h\to 0} \dfrac{h+3}{\sqrt{4+h}(2+h+\sqrt{4+h})} \\
= & \dfrac{3}{\sqrt{4+0}(2+0+\sqrt{4+0})} \\
= & \dfrac{3}{8}
\end{array}$