Question

equaçao da reta tagente a curva y^3+x^2=0 pontos (1,-1)

Answer 1

$y^{3}+x^{2}=0 \\ \\ 3y^{2} \, \displaystyle \frac{dy}{dx}+2x=0 \\ \\ 3y^{2} \, \displaystyle \frac{dy}{dx}=-2x \\ \\ \displaystyle \frac{dy}{dx}= -\frac{2x}{3y^{2}} \\ \\ \\ m= -\frac{2x}{3y^{2}} \\ \\ m=-\frac{2}{3} \\ \\ \\ y-y_o=m(x-x_o) \\ \\ y+1=-\frac{2}{3}(x-1) \\ \\ y=-\frac{2}{3}x-\frac{1}{3} \\ \\ y=\frac{-2x-1}{3} \\ \\ 3y=-2x-1 \\ \\ \boxed{\boxed{2x+3y+1=0}}$