The first question of the second section in the first part of your homework asks to differentiate the function using implicit and logarithmic differentiation:

1. $(x^2 + y^2)^2 = x^2 - y^2$

Let's solve this step by step using both implicit differentiation and logarithmic differentiation.

Implicit Differentiation

Start by differentiating both sides of the equation implicitly with respect to $x$, keeping in mind that $y$ is a function of $x$, so we'll need to use the chain rule for terms involving $y$.

The equation is:

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

1. Differentiate both sides with respect to $x$: $\frac{d}{dx} \left( (x^2 + y^2)^2 \right) = \frac{d}{dx} \left( x^2 - y^2 \right)$

2. For the left-hand side, apply the chain rule: $2(x^2 + y^2) \cdot \frac{d}{dx} (x^2 + y^2)$ Now, differentiate $x^2 + y^2$: $\frac{d}{dx} (x^2 + y^2) = 2x + 2y \frac{dy}{dx}$ Thus, the left-hand side becomes: $2(x^2 + y^2) \cdot (2x + 2y \frac{dy}{dx})$

3. For the right-hand side: $\frac{d}{dx} (x^2 - y^2) = 2x - 2y \frac{dy}{dx}$

Now, equate both sides: $2(x^2 + y^2) \cdot (2x + 2y \frac{dy}{dx}) = 2x - 2y \frac{dy}{dx}$

This is an implicit derivative equation. To solve for $\frac{dy}{dx}$, simplify and collect terms with $\frac{dy}{dx}$.

Logarithmic Differentiation

For logarithmic differentiation, take the natural logarithm of both sides of the equation: $\ln \left( (x^2 + y^2)^2 \right) = \ln(x^2 - y^2)$

1. Use the logarithmic identity $\ln(a^b) = b \ln(a)$: $2 \ln(x^2 + y^2) = \ln(x^2 - y^2)$

2. Differentiate both sides with respect to $x$: $2 \cdot \frac{1}{x^2 + y^2} \cdot (2x + 2y \frac{dy}{dx}) = \frac{1}{x^2 - y^2} \cdot (2x - 2y \frac{dy}{dx})$

Simplify the equation to solve for $\frac{dy}{dx}$.

Would you like me to continue with the algebraic steps to isolate $\frac{dy}{dx}$?