To solve this problem, we'll implicitly differentiate the given equation twice to find $\frac{d^2 y}{dx^2}$.

Step-by-Step Solution

The equation is: $y^2 - xy + x^2 = 4$

Step 1: Differentiate with respect to $x$ (First Derivative)

Since we have an equation involving both $x$ and $y$, we'll use implicit differentiation.

1. Differentiate $y^2$ with respect to $x$: $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx} = 2y \cdot y'$

2. Differentiate $-xy$ with respect to $x$: $\frac{d}{dx}(-xy) = -\left(x \frac{dy}{dx} + y\right) = -\left(xy' + y\right)$

3. Differentiate $x^2$ with respect to $x$: $\frac{d}{dx}(x^2) = 2x$

Since the right-hand side is a constant $4$, its derivative with respect to $x$ is $0$.

Putting it all together: $2y \cdot y' - (xy' + y) + 2x = 0$

Simplify to solve for $y'$: $2y \cdot y' - xy' - y + 2x = 0$ $y'(2y - x) = y - 2x$ $y' = \frac{y - 2x}{2y - x}$

Step 2: Differentiate again to find $\frac{d^2 y}{dx^2}$ (Second Derivative)

Now that we have $y' = \frac{y - 2x}{2y - x}$, we need to differentiate it with respect to $x$ again to find $\frac{d^2 y}{dx^2}$.

Using the quotient rule: $y'' = \frac{(2y - x) \cdot \frac{d}{dx}(y - 2x) - (y - 2x) \cdot \frac{d}{dx}(2y - x)}{(2y - x)^2}$

Differentiate the numerator and denominator:

1. $\frac{d}{dx}(y - 2x) = y' - 2$
2. $\frac{d}{dx}(2y - x) = 2y' - 1$

Substitute and simplify: $y'' = \frac{(2y - x)(y' - 2) - (y - 2x)(2y' - 1)}{(2y - x)^2}$

Substitute $y' = \frac{y - 2x}{2y - x}$ into this expression to get a simplified form.

Would you like me to continue with detailed simplification, or do you need clarification on any of the steps so far?

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Expanding Questions

1. How would this process differ if the original equation were $y^2 - xy + x^3 = 4$?
2. Can you explain why we use the quotient rule in finding the second derivative?
3. How does implicit differentiation apply if we add another variable, say $z$, into the equation?
4. How can we verify our second derivative once found?
5. How would the process change if the equation included trigonometric functions?

Tip

Remember that implicit differentiation often involves using both the product and quotient rules, especially when differentiating the second time.