Answer
The difference quotient for the given function is $\dfrac{1}{\sqrt{x+h}+\sqrt{x}}$
Work Step by Step
$f(x)=\sqrt{x}$
Find the difference quotient $\dfrac{f(x+h)-f(x)}{h}$
Start by finding $f(x+h)$. Substitute $x$ by $x+h$ in $f(x)$ and simplify:
$f(x+h)=\sqrt{x+h}$
Substitute the known values into the formula for the difference quotient:
$\dfrac{f(x+h)-f(x)}{h}=\dfrac{\sqrt{x+h}-\sqrt{x}}{h}=...$
Rationalize the numerator and simplify:
$...=\dfrac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}=\dfrac{(\sqrt{x+h})^{2}-(\sqrt{x})^{2}}{h(\sqrt{x+h}+\sqrt{x})}=...$
$...=\dfrac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}=\dfrac{h}{h(\sqrt{x+h}+\sqrt{x})}=\dfrac{1}{\sqrt{x+h}+\sqrt{x}}$
The difference quotient for the given function is $\dfrac{1}{\sqrt{x+h}+\sqrt{x}}$
