Ответ:

[tex]\dfrac{-22x^2 + 20 x + 7}{(2x^2 - 3x + 2)^2}[/tex]

Пошаговое объяснение:

[tex]\displaystyle y' = \bigg ( \frac{4x^2 + 5x - 1}{2x^2 - 3x + 2} \bigg ) ' = \bigg ( 2+ \frac{11x - 5}{2x^2 - 3x + 2} \bigg ) ' = 2' + \bigg ( \frac{11x - 5}{2x^2 - 3x + 2}\bigg ) ' =\\\\\ \\ = \bigg ( \frac{11x - 5}{2x^2 - 3x + 2}\bigg ) '[/tex]

Производная частного

[tex]\bigg ( \dfrac{u}{v} \bigg ) ' = \dfrac{u'v - uv'}{v^2} \Rightarrow[/tex]

[tex]\displaystyle \bigg ( \frac{11x - 5}{2x^2 - 3x + 2}\bigg ) ' = \frac{(11x - 5)' (2x^2 - 3x + 2) - (11x - 5)(2x^2 - 3x+2 )'}{(2x ^2 -3x + 2)^2} = \\\\\\\ =\frac{11(2x^ 2 - 3x + 2 ) - (11x - 5)(4x - 3) }{(2x^2 - 3x + 2)^2} = \frac{22x^2 -33x + 22 - (44x^2 -53x + 15)}{(2x^2 - 3x + 2)^2} = \\\\\ =\frac{-22x^2 + 20 x + 7}{(2x^2 - 3x + 2)^2}[/tex]