## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\frac{f(x+h)-f(x)}{h}=\frac{2x+h}{\sqrt{(x+h)^2+1}+\sqrt{x^2+1}}.$$
The expression becomes $$\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{h}=\frac{\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{h}\cdot\frac{\sqrt{(x+h)^2+1}+\sqrt{x^2+1}}{\sqrt{(x+h)^2+1}+\sqrt{x^2+1}}=\frac{\sqrt{(x+h)^2+1}^2-\sqrt{x^2+1}^2}{h(\sqrt{(x+h)^2+1}+\sqrt{x^2+1})}=\frac{(x+h)^2+1-x^2-1}{h(\sqrt{(x+h)^2+1}+\sqrt{x^2+1})}=\frac{x^2+2xh+h^2-x^2}{h(\sqrt{(x+h)^2+1}+\sqrt{x^2+1})}=\frac{2xh+h^2}{h(\sqrt{(x+h)^2+1}+\sqrt{x^2+1})}=\frac{2x+h}{\sqrt{(x+h)^2+1}+\sqrt{x^2+1}}.$$