1. **State the problem:** We need to find $\frac{dy}{dx}$ for the equation $$2 y^3 + 2 x^2 = 4$$ using implicit differentiation. 2. **Recall the rule:** When differentiating implicitly, treat $y$ as a function of $x$, so use the chain rule for terms involving $y$. The derivative of $y^n$ with respect to $x$ is $$n y^{n-1} \frac{dy}{dx}$$. 3. **Differentiate both sides with respect to $x$:** $$\frac{d}{dx}(2 y^3) + \frac{d}{dx}(2 x^2) = \frac{d}{dx}(4)$$ 4. **Apply the derivatives:** - For $2 y^3$, use the chain rule: $$2 \cdot 3 y^2 \frac{dy}{dx} = 6 y^2 \frac{dy}{dx}$$ - For $2 x^2$, derivative is: $$2 \cdot 2 x = 4 x$$ - The derivative of constant 4 is 0. So the differentiated equation is: $$6 y^2 \frac{dy}{dx} + 4 x = 0$$ 5. **Solve for $\frac{dy}{dx}$:** $$6 y^2 \frac{dy}{dx} = -4 x$$ Divide both sides by $6 y^2$: $$\frac{dy}{dx} = \frac{-4 x}{6 y^2}$$ Show cancellation: $$\frac{dy}{dx} = \frac{\cancel{2} \cdot -2 x}{\cancel{2} \cdot 3 y^2} = \frac{-2 x}{3 y^2}$$ 6. **Final answer:** $$\boxed{\frac{dy}{dx} = \frac{-2 x}{3 y^2}}$$ This is the derivative of $y$ with respect to $x$ found by implicit differentiation.