Odpowiedź:
Szczegółowe wyjaśnienie:
[tex]f`(x)=(\frac{2-x^ 3}{(x-1)^2} -\frac{2x}{x-1} )`=(\frac{2-x^ 3}{(x-1)^2})` -(\frac{2x}{x-1} )`=[/tex]
[tex]=\frac{(2-x^3)`*(x-1)^2-(2-x^3)[(x-1)^2]`}{[(x-1)^2]^2} -\frac{(2x)`*(x-1)-2x*(x-1)`}{(x-1)^2} =\\[/tex]
[tex]=\frac{-3x^2(x-1)^2-(2-x^3)*2(x-1)*1}{(x-1)^4} -\frac{2(x-1)-2x*1}{(x-1)^2} =\\=\frac{-3x^2(x-1)+2(x^3-2)}{(x-1)^3} -\frac{2x-2-2x}{(x-1)^2} =\frac{-3x^3+3x^2+2x^3-4}{(x-1)^3} -\frac{(x-1)*(-2)}{(x-1)^3} =\frac{-x^3+3x^2-4+2x-2}{(x-1)^3} =\frac{-x^3+3x^2+2x-6}{(x-1)^3} =\frac{-x^2(x-3)+2(x-3)}{(x-1)^3} =\frac{(x-3)(-x^2+2)}{(x-1)^3} =\frac{(x-3)(\sqrt{2}-x)(\sqrt{2}+x) }{(x-1)^3} \\[/tex]