Ответ:

-0,09

Пошаговое объяснение:

[tex]y(x)=\frac{9}{4+x^3}\\\\y`(x)=(\frac{9}{4+x^3})`=\frac{(9)`(4+x^3)-9(4+x^3)`}{(4+x^3)^2}=\frac{0*(4+x^3)-9*(0+3x^2)}{(4+x^3)^2}=\frac{0-9*0-27x^2}{(4+x^3)^2}=\\\\=-\frac{27x^2}{(4+x^3)^2}\\\\\\x=4\\\\y`(x)=y`(4)=-\frac{27*4^2}{(4+4^3)^2}=-\frac{27*16}{(4+64)^2}=-\frac{432}{68^2}=-\frac{432}{4624}\approx-0,093425...=-0,09[/tex]