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 !! 
 A: you did wrong in the first step, the derivate of (x+3) is equal to 1. 
A: The derivative of $2x-1$ is $2$.
The derivative of $(x+3)^{1/2}$ is
$$
\frac{1}{2}(x+3)^{-1/2}\cdot \color{red}{1}=\frac{1}{2(x+3)^{1/2}}
$$
Not $(x+0)$ as you wrote, but this disappeared in the second step.
The product rule gives
$$
2(x+3)^{1/2}+(2x-1)\frac{1}{2(x+3)^{1/2}}
=\frac{4x+12+2x-1}{2(x+3)^{1/2}}
=\frac{6x+11}{2(x+3)^{1/2}}
$$
You divided the numerator by $2$ and forgot to cancel it in the denominator.
A: 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*}
A: Using production rule and rule for derivative of complicated function:
$$(f\cdot g)' = f'\cdot g + f \cdot g' \\ f(\varphi(x))' = f' \cdot (\varphi(x))'$$
We have:
$$(2x-1)'\cdot(x+3)^{\frac{1}{2}}+(2x-1)\cdot((x+3)^{\frac{1}{2}})'\cdot(x+3)'$$
$$(2x-1)' = 2 \\ ((x+3)^{\frac{1}{2}})' = \frac{1}{2}\cdot(x+3)^{\frac{-1}{2}} \\ (x+3)' = 1$$
You can do the rest I think :)
