The problem asks us to differentiate the given equation with respect to $x$. The equation provided is:

$\frac{x + y}{x - y} = x^2 + y^2$

Let us find the derivative step by step:

Step 1: Differentiate both sides of the equation

The left-hand side is a fraction, so we use the quotient rule, and for the right-hand side, we apply the chain rule as needed. Recall that:

$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

Let $u = x + y$ and $v = x - y$. Differentiating with respect to $x$:

Left-hand side:

$\frac{d}{dx} \left(\frac{x + y}{x - y}\right) = \frac{(x - y)\frac{d}{dx}(x + y) - (x + y)\frac{d}{dx}(x - y)}{(x - y)^2}$

Here, derivatives of $x + y$ and $x - y$ are: $\frac{d}{dx}(x + y) = 1 + \frac{dy}{dx}, \quad \frac{d}{dx}(x - y) = 1 - \frac{dy}{dx}$

Substitute into the quotient rule: $\frac{(x - y)(1 + \frac{dy}{dx}) - (x + y)(1 - \frac{dy}{dx})}{(x - y)^2}$

Right-hand side:

Differentiate $x^2 + y^2$: $\frac{d}{dx}(x^2 + y^2) = 2x + 2y \frac{dy}{dx}$

Step 2: Combine results

Equating the derivatives on both sides: $\frac{(x - y)(1 + \frac{dy}{dx}) - (x + y)(1 - \frac{dy}{dx})}{(x - y)^2} = 2x + 2y \frac{dy}{dx}$

Would you like me to simplify further and solve explicitly for $\frac{dy}{dx}$? Let me know!

Related Questions:

1. What is the quotient rule, and how is it applied in differentiation?
2. How do you differentiate implicit functions like $x + y = 0$?
3. What are some practical uses of implicit differentiation in physics?
4. How can we verify if the calculated derivative is correct?
5. How do you interpret $\frac{dy}{dx}$ geometrically?

Tip:

When differentiating equations with fractions, the quotient rule simplifies the process systematically. Always identify numerator and denominator explicitly.