Answer
$$\frac{1}{\sqrt{2 x+1}}$$
Work Step by Step
Given $$y=\sqrt{2 x+1}$$
Now we find the derivative:
\begin{align*}
\frac{d y}{d x}&=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\\
&=\lim _{h \rightarrow 0} \frac{\sqrt{2 x+2 h+1}-\sqrt{2 x+1}}{h}\\
&=\lim _{h \rightarrow 0} \frac{\sqrt{2 x+2 h+1}-\sqrt{2 x+1}}{h} \cdot \frac{\sqrt{2 x+2 h+1}+\sqrt{2 x+1}}{\sqrt{2 x+2 h+1}+\sqrt{2 x+1}}\\
&= \lim _{h \rightarrow 0} \frac{(2 x+2 h+1)-(2 x+1)}{h(\sqrt{2 x+2 h+1}+\sqrt{2 x+1})}\\
&=\lim _{h \rightarrow 0} \frac{2 h}{h(\sqrt{2 x+2 h+1}+\sqrt{2 x+1})}\\
&=\frac{1}{\sqrt{2 x+1}}
\end{align*}
