Answer:
[tex]\dfrac{dy}{dx}=\dfrac{4y-2xy\ln y}{3y^3+x^2}[/tex]
Step-by-step explanation:
[tex]y^3+x^2\ln y=4x+2[/tex]
Diferenciando con respecto a [tex]x[/tex]
[tex]3y^2\dfrac{dy}{dx}+2x\ln y+\dfrac{x^2}{y}\dfrac{dy}{dx}=4\\\Rightarrow (3y^2+\dfrac{x^2}{y})\dfrac{dy}{dx}=4-2x\ln y\\\Rightarrow (\dfrac{3y^3+x^2}{y})\dfrac{dy}{dx}=4-2x\ln y\\\Rightarrow \dfrac{dy}{dx}=\dfrac{4y-2xy\ln y}{3y^3+x^2}[/tex]
La derivada de la función dada es [tex]\dfrac{4y-2xy\ln y}{3y^3+x^2}[/tex].
