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