We are tasked with finding $y''$ by implicit differentiation of the equation:

$x^2 + 2y^2 = 2$

Step 1: Differentiate both sides with respect to $x$

We apply implicit differentiation, remembering that $y$ is a function of $x$ (so we must use the chain rule when differentiating terms involving $y$):

$\frac{d}{dx}(x^2) + \frac{d}{dx}(2y^2) = \frac{d}{dx}(2)$

This becomes:

$2x + 2 \cdot 2y \cdot \frac{dy}{dx} = 0$

Simplifying:

$2x + 4y \frac{dy}{dx} = 0$

Step 2: Solve for $\frac{dy}{dx}$

Rearranging the above equation:

$4y \frac{dy}{dx} = -2x$

Dividing both sides by $4y$:

$\frac{dy}{dx} = -\frac{x}{2y}$

Step 3: Differentiate again to find $y''$

We now differentiate $\frac{dy}{dx} = -\frac{x}{2y}$ implicitly to find $y''$:

$\frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dx} \left( -\frac{x}{2y} \right)$

Using the quotient rule for the right-hand side:

$\frac{d}{dx} \left( -\frac{x}{2y} \right) = - \frac{(2y \cdot 1 - x \cdot 2 \frac{dy}{dx})}{(2y)^2}$

Substitute $\frac{dy}{dx} = -\frac{x}{2y}$:

$y'' = - \frac{2y - x \cdot 2 \cdot \left( -\frac{x}{2y} \right)}{4y^2}$

Simplifying:

$y'' = - \frac{2y + x^2/y}{4y^2}$

Multiply through to clear the fraction inside:

$y'' = - \frac{2y^2 + x^2}{4y^3}$

Thus, the second derivative is:

$y'' = -\frac{2y^2 + x^2}{4y^3}$

Final Answer:

$y'' = -\frac{2y^2 + x^2}{4y^3}$

Would you like more details on any of these steps or have any questions?

Here are 5 related questions to expand on this:

1. How does the chain rule apply in implicit differentiation?
2. What are the key differences between explicit and implicit differentiation?
3. Can you explain the quotient rule used in finding $y''$?
4. Why do we use implicit differentiation when $y$ is a function of $x$?
5. How do we solve higher-order derivatives using implicit differentiation?

Tip: When differentiating implicitly, always apply the chain rule to every term involving the dependent variable.