Answer
$\dfrac{1}{\sqrt {x+h+1}+\sqrt {x+1}}$
Work Step by Step
Step $1$.
Given $f(x)=\sqrt {x+1}$, we have $f(x+h)=\sqrt {x+h+1}$
Step $2$.
$f(x+h)-f(x)=\sqrt {x+h+1}-\sqrt {x+1}$
Step $3$.
(Rationalize the numerator as suggested.)
$\begin{align*}
\dfrac{f(x+h)-f(x)}{h}&=\frac{\sqrt {x+h+1}-\sqrt {x+1}}{h}\\
&=\frac{\sqrt {x+h+1}-\sqrt {x+1}}{h}\cdot\frac{\sqrt {x+h+1}+\sqrt {x+1}}{\sqrt {x+h+1}+\sqrt {x+1}}\\
&=\frac{x+h+1-x-1}{h(\sqrt {x+h+1}+\sqrt {x+1})}\\
&=\frac{h}{h(\sqrt {x+h+1}+\sqrt {x+1})}\\
&=\frac{1}{\sqrt {x+h+1}+\sqrt {x+1}}
\end{align*}$
