Answer:[tex]\frac{dy}{dx}=\frac{-(4x^{3}+2xy^{2})}{2x^{2}y+3y^{2}}[/tex]Explanation:
The given equation is:
[tex]x^4+x^2y^2+y^3=5[/tex]
The differential is given as:
[tex]4x^3+2xy^2+2yx^2\frac{dy}{dx}+3y^2\frac{dy}{dx}=0[/tex]
Make dy/dx the subject of the formula:
[tex]\begin{gathered} 2yx^2\frac{dy}{dx}+3y^2\frac{dy}{dx}=-(4x^3+2xy^2) \\ \\ \frac{dy}{dx}(2x^2y+3y^2)=-(4x^3+2xy^2) \\ \\ \frac{dy}{dx}=\frac{-(4x^3+2xy^2)}{2x^2y+3y^2} \end{gathered}[/tex]
