1. * find dy/dx if x² + 2y³ = x³ - 4y²

Explanation:

Step1: Differentiate both sides with respect to x

Differentiate each term: For $x^2$, using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have $\frac{d}{dx}(x^2)=2x$. For $2y^3$, using the chain - rule $\frac{d}{dx}(f(y)) = f^\prime(y)\frac{dy}{dx}$, we get $2\times3y^{2}\frac{dy}{dx}=6y^{2}\frac{dy}{dx}$. For $x^3$, $\frac{d}{dx}(x^3)=3x^{2}$. For $-4y^2$, using the chain - rule, we have $-4\times2y\frac{dy}{dx}=-8y\frac{dy}{dx}$. So, $2x + 6y^{2}\frac{dy}{dx}=3x^{2}-8y\frac{dy}{dx}$.

Step2: Isolate $\frac{dy}{dx}$ terms

Move all terms with $\frac{dy}{dx}$ to one side: $6y^{2}\frac{dy}{dx}+8y\frac{dy}{dx}=3x^{2}-2x$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(6y^{2}+8y)=3x^{2}-2x$.

Step3: Solve for $\frac{dy}{dx}$

Divide both sides by $6y^{2}+8y$: $\frac{dy}{dx}=\frac{3x^{2}-2x}{6y^{2}+8y}$.

Answer:

$\frac{3x^{2}-2x}{6y^{2}+8y}$