## Precalculus: Mathematics for Calculus, 7th Edition

$\dfrac{\sqrt{x+h}-\sqrt{x}}{h}=\dfrac{1}{\sqrt{x+h}+\sqrt{x}}$
$\dfrac{\sqrt{x+h}-\sqrt{x}}{h}$ (rationalize the numerator) Multiply the numerator and the denominator of the rational expression by the conjugate of the numerator and simplify: $\dfrac{\sqrt{x+h}-\sqrt{x}}{h}=\dfrac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}=...$ $...=\dfrac{(\sqrt{x+h})^{2}-(\sqrt{x})^{2}}{h(\sqrt{x+h}+\sqrt{x})}=\dfrac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}=...$ $...=\dfrac{h}{h(\sqrt{x+h}+\sqrt{x})}=\dfrac{1}{\sqrt{x+h}+\sqrt{x}}$