Answer
$f'(x)=\frac{(x-1)(x+3)+8x+4\sqrt x}{(x+3)^{2}\sqrt x}$
Work Step by Step
We find:
$f'(x)=\frac{[(2\sqrt x+1)(x-1)]'(x+3)-(2\sqrt x+1)(x-1)(x+3)'}{(x+3)^{2}}$
$=\frac{[(2\sqrt x+1)'(x-1)+(2\sqrt x+1)(x-1)'](x+3)-(2\sqrt x+1)(x-1)}{(x+3)^{2}}$
$=\frac{\frac{(x-1)(x+3)}{\sqrt x}+(2\sqrt x+1)(x+3)-(2\sqrt x+1)(x+1)}{(x+3)^{2}}$
$=\frac{\frac{(x-1)(x+3)}{\sqrt x}-(2\sqrt x+1)(x-1-x-3)}{(x+3)^{2}}$
$=\frac{\frac{(x-1)(x+3)}{\sqrt x}+\frac{4(2\sqrt x+1)\sqrt x}{\sqrt x}}{(x+3)^{2}}$
$=\frac{(x-1)(x+3)+8x+4\sqrt x}{(x+3)^{2}\sqrt x}$
