Spiegare gentilmente i ragionamenti, i passaggi e argomentare.
$ y(x) = \sqrt{\frac{1-x}{x+3}} $
$ y'(x) = \frac{1}{2} \sqrt{\frac{x+3}{1-x}} \; \; (\frac{1-x}{x+3})' $
$ = \frac{1}{2} \sqrt{\frac{x+3}{1-x}} \frac{-x-3-1+x}{(x+3)^2} $
$ = \frac{1}{2} \sqrt{\frac{x+3}{1-x}} \frac{-4}{(x+3)^2} $
$ = \sqrt{\frac{x+3}{1-x}} \cdot \frac{-2}{(x+3)^2} $
$ = \frac{-2}{(x+3)^2} \sqrt{\frac{x+3}{1-x}}$