Let's find the first derivative of the equation:

Given:

$x^2(x - y)^2 = x^2 - y^2$

Differentiate both sides with respect to x:

Step 1: Differentiate Left-Hand Side

Apply the product rule:
[ \frac{d}{dx}[x^2(x - y)^2] = \frac{d}{dx}(x^2) \cdot (x - y)^2 + x^2 \cdot \frac{d}{dx}[(x - y)^2] ]

• Derivative of $x^2$: $2x(x - y)^2$

• Derivative of $(x - y)^2$ using the chain rule: $2(x - y) \cdot (1 - y')$

So the whole left-hand side becomes: $2x(x - y)^2 + x^2 \cdot 2(x - y)(1 - y')$

Step 2: Differentiate Right-Hand Side

$\frac{d}{dx}(x^2 - y^2) = 2x - 2y \cdot y'$

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Final Equation:

Now equate both sides: $2x(x - y)^2 + 2x^2(x - y)(1 - y') = 2x - 2y \cdot y'$

Simplify this and solve for $y'$.

Would you like me to simplify the entire equation and solve explicitly for $y'$?