Question Detail

To find $\frac{dy}{dx}$ for the implicit function $3x^2 y + xy^2 = 5$, we differentiate each term with respect to $x$.$d/dx(3x^2 y) + d/dx(xy^2) = d/dx(5)$Using the product rule:$d/dx(3x^2 y) = (d/dx(3x^2))y + 3x^2(d/dx(y)) = 6xy + 3x^2 \frac{dy}{dx}$$d/dx(xy^2) = (d/dx(x))y^2 + x(d/dx(y^2)) = 1 \cdot y^2 + x(2y \frac{dy}{dx}) = y^2 + 2xy \frac{dy}{dx}$$d/dx(5) = 0$Substitute these back into the equation:$(6xy + 3x^2 \frac{dy}{dx}) + (y^2 + 2xy \frac{dy}{dx}) = 0$Group terms with $\frac{dy}{dx}$:$3x^2 \frac{dy}{dx} + 2xy \frac{dy}{dx} = -6xy - y^2$$\frac{dy}{dx} (3x^2 + 2xy) = -(6xy + y^2)$Solve for $\frac{dy}{dx}$:$\frac{dy}{dx} = -\frac{y^2 + 6xy}{3x^2 + 2xy}$