\( z = \ln \frac { x ^ { 2 } - y ^ { 2 } } { x ^ { 2 } + y ^ { 2 } } \)

The expression you've provided is: \[ z = \ln \frac{x^2 - y^2}{x^2 + y^2} \] This is a logarithmic function that takes the ratio of two quadratic expressions involving \(x\) and \(y\). Let's analyze this expression further. ### Domain For the logarithm to be defined, the argument must be positive: \[ \frac{x^2 - y^2}{x^2 + y^2} > 0 \] This implies two conditions: 1. \(x^2 - y^2 > 0\) (the numerator must be positive) 2. \(x^2 + y^2 > 0\) (the denominator must be positive) The second condition is always true for real \(x\) and \(y\) unless both \(x\) and \(y\) are zero, which would make the expression undefined. The first condition simplifies to: \[ x^2 > y^2 \quad \Rightarrow \quad |x| > |y| \] Thus, the domain of \(z\) is where \(|x| > |y|\). ### Behavior - As \(x\) approaches \(y\) from either side, \(z\) approaches \(-\infty\) because the numerator approaches zero. - As \(|x|\) becomes much larger than \(|y|\), \(z\) approaches \(0\) since the ratio approaches \(1\). ### Partial Derivatives If you are interested in the behavior of \(z\) with respect to \(x\) and \(y\), we can compute the partial derivatives: 1. **Partial derivative with respect to \(x\)**: Using the quotient rule and chain rule: \[ \frac{\partial z}{\partial x} = \frac{1}{\frac{x^2 - y^2}{x^2 + y^2}} \cdot \frac{\partial}{\partial x} \left( \frac{x^2 - y^2}{x^2 + y^2} \right) \] Calculating the derivative of the fraction: \[ \frac{\partial}{\partial x} \left( \frac{x^2 - y^2}{x^2 + y^2} \right) = \frac{(2x)(x^2 + y^2) - (x^2 - y^2)(2x)}{(x^2 + y^2)^2} \] This simplifies to: \[ \frac{2x(y^2 + y^2)}{(x^2 + y^2)^2} = \frac{4xy^2}{(x^2 + y^2)^2} \] Thus, \[ \frac{\partial z}{\partial x} = \frac{4xy^2}{(x^2 + y^2)(x^2 - y^2)} \] 2. **Partial derivative with respect to \(y\)**: Similarly, we can find: \[ \frac{\partial z}{\partial y} = \frac{1}{\frac{x^2 - y^2}{x^2 + y^2}} \cdot \frac{\partial}{\partial y} \left( \frac{x^2 - y^2}{x^2 + y^2} \right) \] Calculating the derivative of the fraction: \[ \frac{\partial}{\partial y} \left( \frac{x^2 - y^2}{x^2 + y^2} \right) = \frac{(-2y)(x^2 + y^2) - (x^2 - y^2)(2y)}{(x^2 + y^2)^2} \] This simplifies to: \[ \frac{-2y(x^2 + y^2 + x^2 - y^2)}{(x^2 + y^2)^2} = \frac{-4yx^2}{(x^2 + y^2)^2} \] Thus, \[ \frac{\partial z}{\partial y} = \frac{-4yx^2}{(x^2 + y^2)(x^2 - y^2)} \] ### Summary The expression \( z = \ln \frac{x^2 - y^2}{x^2 + y^2} \) is defined for \(|x| > |y|\) and approaches \(-\infty\) as \(x\) approaches \(y\) and \(0\) as \(|x|\) becomes much larger than \(|y|\). The partial derivatives provide insight into how \(z\) changes with