## Calculus: Early Transcendentals (2nd Edition)

The soluton is $$\frac{f(x+h)-f(x)}{h}=3\frac{1}{\sqrt{x(x+h)}(\sqrt{x+h}+\sqrt{x})}.$$
The expression becomes $$\frac{f(x+h)-f(x)}{h}=\frac{-\frac{3}{\sqrt{x+h}}-\left(-\frac{3}{\sqrt{x}}\right)}{h}=3\frac{\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+h}}}{h}=3\frac{\frac{\sqrt{x+h}-\sqrt{x}}{\sqrt{x(x+h)}}}{h}=3\frac{\sqrt{x+h}-\sqrt{x}}{h\sqrt{x(x+h)}}=3\frac{\sqrt{x+h}-\sqrt{x}}{h\sqrt{x(x+h)}}\cdot\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}=3\frac{\sqrt{x+h}^2-\sqrt{x}^2}{h\sqrt{x(x+h)}(\sqrt{x+h}+\sqrt{x})}=3\frac{x+h-x}{h\sqrt{x(x+h)}(\sqrt{x+h}+\sqrt{x})}=3\frac{h}{h\sqrt{x(x+h)}(\sqrt{x+h}+\sqrt{x})}=3\frac{1}{\sqrt{x(x+h)}(\sqrt{x+h}+\sqrt{x})}.$$