Respuesta :

nobillionaireNobley
[tex]y= \sqrt[4]{(2x^3-1)} \\ y^4=2x^3-1 \\ 2x^3=y^4+1 \\ x^3= \frac{y^4+1}{2} \\ x= \sqrt[3]{\frac{y^4+1}{2}} [/tex]

Therefore, F'(x) = [tex]\sqrt[3]{\frac{y^4+1}{2}}[/tex]