Answer
The difference quotient for the provided function is \[\frac{1}{\sqrt{x+h}+\sqrt{x}}\].
Work Step by Step
Consider the provided function: $f\left( x \right)=\sqrt{x}$.
Now, substitute $x=x+h$ in the above equation to find $f\left( x+h \right)$
That is,
$f\left( x+h \right)=\sqrt{x+h}$
Now, apply the difference quotient formula,
$\begin{align}
& \frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{\left( \sqrt{x+h}-\sqrt{x} \right)\left( \sqrt{x+h}+\sqrt{x} \right)}{h\left( \sqrt{x+h}+\sqrt{x} \right)} \\
& =\frac{{{\left( \sqrt{x+h} \right)}^{2}}-{{\left( \sqrt{x} \right)}^{2}}}{h\left( \sqrt{x+h}+\sqrt{x} \right)} \\
& =\frac{x+h-x}{h\left( \sqrt{x+h}+\sqrt{x} \right)} \\
& =\frac{1}{\sqrt{x+h}+\sqrt{x}}
\end{align}$
Hence, the difference quotient for the provided function is $\frac{1}{\sqrt{x+h}+\sqrt{x}}$.
