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 $x$.