Ответ:

[tex]\dfrac{x^4 + 3x^2 }{(x^2 +1)^2}[/tex]

Объяснение:

Производная частного

[tex]\bullet ~~\displaystyle \bigg ( \frac{u}{v } \bigg ) ' = \frac{u'v - uv'}{v^2}[/tex]

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