To find $y''$ using implicit differentiation, we first start with the given equation:

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

Step 1: Differentiate implicitly with respect to $x$

We apply implicit differentiation to both sides of the equation. Using the chain rule for $y^2$, we treat $y$ as a function of $x$.

1. Differentiate $x^2$: $\frac{d}{dx}(x^2) = 2x$
2. Differentiate $2y^2$: $\frac{d}{dx}(2y^2) = 2 \cdot 2y \cdot \frac{dy}{dx} = 4y \cdot y'$
3. The derivative of the constant $2$ is 0: $\frac{d}{dx}(2) = 0$

So, after differentiating both sides, we have: $2x + 4y \cdot y' = 0$

Step 2: Solve for $y'$ (first derivative)

Rearranging the equation to solve for $y'$:

$4y \cdot y' = -2x$ $y' = \frac{-2x}{4y} = \frac{-x}{2y}$

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

Now, we differentiate $y' = \frac{-x}{2y}$ to find $y''$. We'll need to use the quotient rule for differentiation: $\frac{d}{dx}\left( \frac{-x}{2y} \right) = \frac{(2y) \cdot (-1) - (-x) \cdot \frac{d}{dx}(2y)}{(2y)^2}$

First, differentiate $2y$ with respect to $x$: $\frac{d}{dx}(2y) = 2 \cdot y'$

Substitute into the quotient rule: $y'' = \frac{(2y)(-1) - (-x)(2 \cdot y')}{(2y)^2}$ $y'' = \frac{-2y + 2x \cdot y'}{4y^2}$

Step 4: Substitute $y' = \frac{-x}{2y}$

Substitute $y' = \frac{-x}{2y}$ into the equation for $y''$: $y'' = \frac{-2y + 2x \cdot \frac{-x}{2y}}{4y^2}$ Simplify: $y'' = \frac{-2y - \frac{x^2}{y}}{4y^2}$ $y'' = \frac{-2y^2 - x^2}{4y^3}$

Final Answer:

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

Since $x^2 + 2y^2 = 2$ from the original equation, we can substitute this into the expression for $y''$: $y'' = \frac{-2}{4y^3} = \frac{-1}{2y^3}$

Thus, the second derivative is: $y'' = \frac{-1}{2y^3}$

Let me know if you want further details or have any questions!

Here are 5 related questions:

1. How does implicit differentiation differ from explicit differentiation?
2. Can implicit differentiation be applied to equations with more complex functions of $y$?
3. What is the significance of using the quotient rule in this problem?
4. Why is it necessary to differentiate both sides of the equation equally when applying implicit differentiation?
5. How would you find the slope of a curve at a specific point using the first derivative $y'$?

Tip: Always remember to apply the chain rule when differentiating terms involving $y$, as it's implicitly a function of $x$.