Derivative of $(2x-1)(x+3)^{\frac{1}{2}}$ Find the derivative of 
$(2x-1)(x+3)^{\frac{1}{2}}$ 
My try - 
$(2x-1)(\frac{1}{2} (x+3)^{\frac{-1}{2}} (x+0) + (x+3)^{\frac{1}{2}} (2)$ 
$ = (x+3)^{\frac{-1}{2}} (\frac{1}{2}(2x-1) + 2(x+3) $ 
$= \frac{3x+5.5}{2 (x+3)^{\frac{1}{2}}} $ 
My numerator is wrong and should be 
$6x+11$ . Where did I go wrong ? Thanks !! 

There are a few issues here - some stray parentheses and an incorrectly written derivative which became corrected on the next step.  I'd proceed like this:
\begin{align*}
  \frac d{dx} (2x-1)(x+3)^{1/2} &= \left[\frac d{dx} (2x-1)\right] \cdot (x+3)^{1/2} + (2x-1) \cdot \frac d{dx} (x+3)^{1/2}\\[0.3cm]
    &= 2 (x+3)^{1/2} + (2x-1) \cdot \frac12(x+3)^{-1/2} \cdot \frac d{dx}(x+3)\\[0.3cm]
    &= 2 (x+3)^{1/2} + \frac{2x-1}{2(x+3)^{1/2}}\\[0.3cm]
    &= 2 (x+3)^{1/2}\cdot \color{red}{\frac{2(x+3)^{1/2}}{2(x+3)^{1/2}}} + \frac{2x-1}{2(x+3)^{1/2}}\\[0.3cm]
    &= \frac{4(x+3) + 2x-1}{2\sqrt{x+3}}\\[0.3cm]
    &= \frac{6x+11}{2\sqrt{x+3}}
\end{align*}
