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The image shows a step-by-step differentiation process of the given equation using implicit differentiation. Let's walk through the steps:

### Given Equation:
\[
(x^2 + y^2)^2 = 2(x^2 - y^2)
\]

### Step 1: Apply Implicit Differentiation
Both sides of the equation are implicitly differentiated with respect to \(x\).

#### Left-hand side:
\[
2(x^2 + y^2) \cdot \frac{d}{dx}(x^2 + y^2)
\]

The chain rule is applied here, where \( (x^2 + y^2)^2 \) is first differentiated as \( 2(x^2 + y^2) \), and then \( (x^2 + y^2) \) is differentiated. The derivative of \( x^2 + y^2 \) is:

\[
\frac{d}{dx}(x^2 + y^2) = 2x + 2y\frac{dy}{dx}
\]

So, the left-hand side becomes:
\[
2(x^2 + y^2)(2x + 2y\frac{dy}{dx})
\]

#### Right-hand side:
\[
\frac{d}{dx}(2(x^2 - y^2)) = 2(2x - 2y\frac{dy}{dx})
\]

### Step 2: Combine and Simplify
Now we equate the derivatives of both sides:
\[
2(x^2 + y^2)(2x + 2y\frac{dy}{dx}) = 2(2x - 2y\frac{dy}{dx})
\]

Divide both sides by 2:
\[
(x^2 + y^2)(2x + 2y\frac{dy}{dx}) = 2x - 2y\frac{dy}{dx}
\]

### Step 3: Expand the Equation
Expand both sides:
\[
2x(x^2 + y^2) + 2y(x^2 + y^2)\frac{dy}{dx} = 2x - 2y\frac{dy}{dx}
\]

### Step 4: Collect \( \frac{dy}{dx} \) terms
Move all terms with \( \frac{dy}{dx} \) to one side:
\[
2y(x^2 + y^2)\frac{dy}{dx} + 2y\frac{dy}{dx} = 2x - 2x(x^2 + y^2)
\]

Factor out \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} [2y(x^2 + y^2) + 2y] = 2x(1 - x^2 - y^2)
\]

### Step 5: Solve for \( \frac{dy}{dx} \)
\[
\frac{dy}{dx} = \frac{2x(1 - x^2 - y^2)}{2y(x^2 + y^2 + 1)}
\]

This is the expression for \( \frac{dy}{dx} \).

### Final Part:
The question at the end asks about the condition under which \( \frac{dy}{dx} = 0 \). This happens when the numerator is zero:
\[
2x(1 - x^2 - y^2) = 0
\]

Thus, \( x = 0 \) or \( x^2 + y^2 = 1 \).

Would you like more details, or do you have any questions about these steps?

### 5 Related Questions:
1. What is implicit differentiation and how does it differ from regular differentiation?
2. How does the chain rule apply when differentiating expressions involving both \( x \) and \( y \)?
3. What happens to the derivative when \( x^2 + y^2 = 1 \)?
4. Why is it important to collect like terms when solving for \( \frac{dy}{dx} \)?
5. How does setting \( \frac{dy}{dx} = 0 \) help find critical points in the function?

### Tip:
When dealing with implicit differentiation, always remember to apply the chain rule carefully, especially when differentiating terms involving \( y \), as they implicitly depend on \( x \).

(x^2 + y^2)^2 = 2(x^2 - y^2), find dy/dx.

Learn how to apply implicit differentiation to the equation (x^2 + y^2)^2 = 2(x^2 - y^2) to find dy/dx. This detailed solution explains the chain rule and algebraic steps necessary to arrive at the correct derivative expression. Suitable for students studying undergraduate calculus.

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