To find $\frac{d^2y}{dx^2}$ for the equation $2y^2 - xy + x^2 = 8$, we first implicitly differentiate the equation with respect to $x$.

Step 1: Differentiate implicitly (first derivative)

The original equation is: $2y^2 - xy + x^2 = 8$

Differentiating each term:

1. $\frac{d}{dx}(2y^2) = 4y \frac{dy}{dx}$ (using the chain rule)
2. $\frac{d}{dx}(-xy) = -x \frac{dy}{dx} - y$ (using the product rule)
3. $\frac{d}{dx}(x^2) = 2x$

Thus, differentiating both sides with respect to $x$: $4y \frac{dy}{dx} - x \frac{dy}{dx} - y + 2x = 0$

Simplify: $(4y - x) \frac{dy}{dx} = y - 2x$

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

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Step 2: Differentiate again (second derivative)

To find $\frac{d^2y}{dx^2}$, differentiate $\frac{dy}{dx} = \frac{y - 2x}{4y - x}$ with respect to $x$ using the quotient rule.

Using the quotient rule:

If $u = y - 2x$ and $v = 4y - x$, then: $\frac{d^2y}{dx^2} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

Step 2.1: Differentiate $u = y - 2x$

$\frac{du}{dx} = \frac{dy}{dx} - 2$

Step 2.2: Differentiate $v = 4y - x$

$\frac{dv}{dx} = 4\frac{dy}{dx} - 1$

Step 2.3: Substitute $\frac{dy}{dx} = \frac{y - 2x}{4y - x}$ into $\frac{du}{dx}$ and $\frac{dv}{dx}$:

1. Substitute $\frac{dy}{dx}$ into $\frac{du}{dx}$: $\frac{du}{dx} = \frac{y - 2x}{4y - x} - 2$

2. Substitute $\frac{dy}{dx}$ into $\frac{dv}{dx}$: $\frac{dv}{dx} = 4\left(\frac{y - 2x}{4y - x}\right) - 1$

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Step 3: Combine into the formula for $\frac{d^2y}{dx^2}$:

Substitute $\frac{du}{dx}$ and $\frac{dv}{dx}$ into the quotient rule formula: $\frac{d^2y}{dx^2} = \frac{(4y - x)\left(\frac{y - 2x}{4y - x} - 2\right) - (y - 2x)\left(4\frac{y - 2x}{4y - x} - 1\right)}{(4y - x)^2}$

This is the general form of the second derivative. Simplify further as needed.

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Want to simplify further or need clarification?

Here are 5 related questions for deeper understanding:

1. How can the chain rule and product rule simplify implicit differentiation?
2. Why is the quotient rule useful in finding $\frac{d^2y}{dx^2}$?
3. How does substituting $\frac{dy}{dx}$ into the second derivative help in implicit differentiation?
4. What are the steps for verifying the correctness of implicit differentiation results?
5. Can this method apply to equations with higher powers of $x$ and $y$?

Tip: Always keep intermediate results organized during implicit differentiation to avoid errors!