Resposta :
[tex]y=\sqrt[3]{\frac{-x^3+x^2+5x-6}{4x}}=(\frac{-x^3+x^2+5x-6}{4x})^{\frac{1}{3}}[/tex]
[tex]y'= \frac{1}{3}*(\frac{-x^3+x^2+5x-6}{4x})^{-\frac{2}{3}}*(-3x^2+2x+5)=\frac{1}{3}*\frac{-3x^2+2x+5}{ \sqrt[3]{ (\frac{-x^3+x^2+5x-6}{4x})^2}} [/tex]
[tex]\frac{1}{3}*\frac{-3x^2+2x+5}{ \frac{\sqrt[3]{(-x^3+x^2+5x-6)^2}}{\sqrt[3]{(4x)^2}}} =\frac{1}{3}*\frac{(-3x^2+2x+5)*\sqrt[3]{16x^2}}{\sqrt[3]{(-x^3+x^2+5x-6)^2}}[/tex]
Hugs
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
[tex]4y^3+12xy^2 \frac{dy}{dx} =-3x^2+2x+5[/tex]
[tex]\frac{dy}{dx} =\frac{-3x^2+2x+5-4y^3}{12xy^2}[/tex]
[tex]y'=\frac{dy}{dx}[/tex]
[tex]y' =\frac{-3x^2+2x+5-4y^3}{12xy^2}[/tex]
[tex]y'= \frac{1}{3}*(\frac{-x^3+x^2+5x-6}{4x})^{-\frac{2}{3}}*(-3x^2+2x+5)=\frac{1}{3}*\frac{-3x^2+2x+5}{ \sqrt[3]{ (\frac{-x^3+x^2+5x-6}{4x})^2}} [/tex]
[tex]\frac{1}{3}*\frac{-3x^2+2x+5}{ \frac{\sqrt[3]{(-x^3+x^2+5x-6)^2}}{\sqrt[3]{(4x)^2}}} =\frac{1}{3}*\frac{(-3x^2+2x+5)*\sqrt[3]{16x^2}}{\sqrt[3]{(-x^3+x^2+5x-6)^2}}[/tex]
Hugs
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
[tex]4y^3+12xy^2 \frac{dy}{dx} =-3x^2+2x+5[/tex]
[tex]\frac{dy}{dx} =\frac{-3x^2+2x+5-4y^3}{12xy^2}[/tex]
[tex]y'=\frac{dy}{dx}[/tex]
[tex]y' =\frac{-3x^2+2x+5-4y^3}{12xy^2}[/tex]