Найти производную функции

$f(x)=\frac{4-x^2}{3+2x}\\ \\ f'(x) = \frac{(4-x^2)'*(3+2x)-(4-x^2)(3+2x)'}{(3+2x)^2} = \frac{-2x(3+2x)-2(4-x^2)}{(3+2x)^2} = \\ \\ = -2\frac{3x+2x^2+4-x^2}{(3+2x)^2} = \frac{-2x^2-6x-8}{4x^2+12x+9} = -\frac{2}{4}*\frac{x^2+3x+4}{x^2+3x+\frac{9}{4}} = \\ \\ = -\frac{1}{2}*(1+\frac{\frac{7}{4}}{(x+\frac{3}{2})^2}) = -\frac{1}{2}(1+\frac{7}{4}*\frac{1}{(x+\frac{3}{2})^2})$

Ответ: $f'(x) = -\frac{1}{2}(1+\frac{7}{4}*\frac{1}{(x+\frac{3}{2})^2})$