3. find the derivative with respect to x. y³ - 4y = 3x² + 2x solve for dy/dx

Explanation:

Step1: Differentiate both sides

Differentiate $y^{3}-4y$ and $3x^{2}+2x$ with respect to $x$. For the left - hand side, use the chain rule. The derivative of $y^{3}$ with respect to $x$ is $3y^{2}\frac{dy}{dx}$, and the derivative of $-4y$ with respect to $x$ is $-4\frac{dy}{dx}$. The derivative of $3x^{2}+2x$ with respect to $x$ is $6x + 2$. So we have $3y^{2}\frac{dy}{dx}-4\frac{dy}{dx}=6x + 2$.

Step2: Factor out $\frac{dy}{dx}$

Factor out $\frac{dy}{dx}$ on the left - hand side: $\frac{dy}{dx}(3y^{2}-4)=6x + 2$.

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

Divide both sides by $3y^{2}-4$ to get $\frac{dy}{dx}=\frac{6x + 2}{3y^{2}-4}$.

Answer:

$\frac{dy}{dx}=\frac{6x + 2}{3y^{2}-4}$