<s>[INST] 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 !! 
 [/INST] 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.
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