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