The image contains a question asking to find the derivative $\frac{dy}{dx}$ using implicit differentiation for the equation:
$2y^3 + \cos y = x^2.$

Solution:

We will differentiate both sides of the equation implicitly with respect to $x$.

Step 1: Write the equation.

$2y^3 + \cos y = x^2.$

Step 2: Differentiate both sides.

Using implicit differentiation, we apply the chain rule where necessary:

1. Derivative of $2y^3$:
   $\frac{d}{dx}(2y^3) = 6y^2 \frac{dy}{dx}.$

2. Derivative of $\cos y$:
   $\frac{d}{dx}(\cos y) = -\sin y \frac{dy}{dx}.$

3. Derivative of $x^2$:
   $\frac{d}{dx}(x^2) = 2x.$

So, the equation becomes: $6y^2 \frac{dy}{dx} - \sin y \frac{dy}{dx} = 2x.$

Step 3: Factor $\frac{dy}{dx}$.

Combine terms involving $\frac{dy}{dx}$: $\frac{dy}{dx}(6y^2 - \sin y) = 2x.$

Step 4: Solve for $\frac{dy}{dx}$.

Divide through by $6y^2 - \sin y$ (assuming $6y^2 - \sin y \neq 0$): $\frac{dy}{dx} = \frac{2x}{6y^2 - \sin y}.$

Final Answer:

$\frac{dy}{dx} = \frac{2x}{6y^2 - \sin y}.$

Would you like a detailed explanation of any step?
Here are 5 related questions to explore this topic further:

1. How is implicit differentiation different from explicit differentiation?
2. What are common mistakes when using the chain rule in implicit differentiation?
3. Can you use the same method to solve $\sin(y) + y^2 = x$?
4. What happens if $6y^2 - \sin y = 0$?
5. How do you verify if your implicit derivative result is correct?

Tip:

When solving implicit differentiation problems, always carefully apply the chain rule to terms involving $y$, and group all $\frac{dy}{dx}$ terms together for easy factoring.