Calculus (3rd Edition)

$$\frac{ -1}{10\sqrt{5}}$$
Since $f(x)=\frac{1}{\sqrt{x}}$, we have $$\frac{f(5+h)-f(5)}{h}=\frac{\frac{1}{\sqrt{5+h}}-\frac{1}{\sqrt{5}}}{h}=\frac{ \sqrt{5}-\sqrt{5+h}}{h\sqrt{5}\sqrt{5+h}}.$$ Now, rationlizing the above fraction by multiplying by $\frac{\sqrt{5}+\sqrt{5+h}}{\sqrt{5}+\sqrt{5+h}}$, we get $$\frac{ \sqrt{5}-\sqrt{5+h}}{h\sqrt{5}\sqrt{5+h}} \frac{\sqrt{5}+\sqrt{5+h}}{\sqrt{5}+\sqrt{5+h}}=\frac{ 5-(5+h)}{h\sqrt{5}\sqrt{5+h}(\sqrt{5}+\sqrt{5+h})} .$$ Finally, we get $$\frac{f(5+h)-f(5)}{h}=-\frac{ 1}{\sqrt{5}\sqrt{5+h}(\sqrt{5}+\sqrt{5+h})} .$$ Plugging in $h=0$, we get: $$-\frac{ 1}{\sqrt{5}\sqrt{5}(\sqrt{5}+\sqrt{5})} =\frac{ -1}{5(2\sqrt{5})}$$ $$=\frac{ -1}{10\sqrt{5}}$$