Answer
$f´(x)=\frac{1}{(x^{2}+3x)}(\frac{2-x)}{\sqrt x}+\frac{(2\sqrt x+1)(x^{2}+4x-6)}{(x^{2}+3x)})$
Work Step by Step
Use product rule:
$f'(x)=(2\sqrt x+1)'(\frac{2-x}{x^{2}+3x})+(2\sqrt x+1)(\frac{2-x}{x^{2}+3x})'$
$=\frac{1}{\sqrt x}(\frac{2-x}{x^{2}+3x})+(2\sqrt x+1)(\frac{(2-x)'(x^{2}+3x-(2-x))(x^{2}+3x)'}{(x^{2}+3x)^{2}})$
$=\frac{1}{(x^{2}+3x)}(\frac{(2-x)}{\sqrt x}+(2\sqrt x+1)(\frac{-x^{2}-3x-4x+2x^{2}+3x-6}{(x^{2}+3x)})$
$=\frac{1}{(x^{2}+3x)}(\frac{2-x}{\sqrt x}+\frac{(2\sqrt x+1)(x^{2}+4x-6)}{(x^{2}+3x)})$
