The problem asks to find the second derivative $\frac{d^2 y}{dx^2}$ from the parameterized equations: $x = \frac{t - 1}{t + 1}, \quad y = \frac{t + 1}{t - 1}$

Steps to solve:

1. First, find $\frac{dx}{dt}$: $x = \frac{t - 1}{t + 1}$ Apply the quotient rule: $\frac{dx}{dt} = \frac{(t + 1)(1) - (t - 1)(1)}{(t + 1)^2} = \frac{(t + 1) - (t - 1)}{(t + 1)^2} = \frac{2}{(t + 1)^2}$

2. Find $\frac{dy}{dt}$: $y = \frac{t + 1}{t - 1}$ Again, apply the quotient rule: $\frac{dy}{dt} = \frac{(t - 1)(1) - (t + 1)(1)}{(t - 1)^2} = \frac{(t - 1) - (t + 1)}{(t - 1)^2} = \frac{-2}{(t - 1)^2}$

3. Find $\frac{dy}{dx}$: Now, we use the chain rule: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-2 / (t - 1)^2}{2 / (t + 1)^2} = -\frac{(t + 1)^2}{(t - 1)^2}$

4. Find $\frac{d^2y}{dx^2}$: Differentiate $\frac{dy}{dx}$ with respect to $t$ again, and use the chain rule. We need $\frac{d}{dt} \left( -\frac{(t + 1)^2}{(t - 1)^2} \right)$, and then divide by $\frac{dx}{dt}$.

Let me compute the second derivative explicitly.The first derivative $\frac{dy}{dx}$ simplifies to:

$\frac{dy}{dx} = -\frac{(t + 1)^2}{(t - 1)^2}$

The second derivative $\frac{d^2y}{dx^2}$ is:

$\frac{d^2 y}{dx^2} = \frac{2(t + 1)^3}{(t - 1)^3}$

This is the required second derivative based on the parameterized equations.

Would you like more details on any step or further clarifications?

Related Questions:

1. How does the quotient rule apply to the differentiation of rational functions?
2. Can you explain how the chain rule works in parametric differentiation?
3. What are the geometric interpretations of first and second derivatives in parametric equations?
4. How do we handle higher-order derivatives in parametric forms?
5. How does the behavior of the second derivative inform us about the curvature of the graph?

Tip: When differentiating parametric equations, always be mindful of applying the chain rule carefully!