Find the derivative respect to $x$? Question
$\sqrt{2x+1}$ 
$=\sqrt{2x+1}$ 
$=\sqrt{2x} + \sqrt{1}$ 
$=\dfrac{1}{2x^{1/2}}$
however the right answer is $\dfrac{1}{\sqrt{2x+1}}$
Can you please help me out? 
this chapter name is (differentiationg rational power
$x^{p/q}$)

Let $y = \sqrt{2x+1}$.
The problem is to find $\dfrac{d}{dx} \sqrt{2x+1}$, i.e. to find $\dfrac{dy}{dx}$.
Let $u=2x+1$.   Then $y=\sqrt{u}$.
Then we have
$$
\frac{dy}{du} = \frac{1}{2\sqrt{u}},\qquad \text{and}\qquad \frac{du}{dx} = 2.
$$
Therefore
$$
\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} = \frac{1}{2\sqrt{u}}\cdot 2 = \frac{1}{\sqrt{u}} = \frac{1}{\sqrt{2x+1}}.
$$
