Single Variable Calculus, Chapter 3, Review Exercises, Section Review Exercises, Problem 48

Determine the equation of the tangent line to the curve $\displaystyle y = \frac{x^2-1}{x^2+1}$ at the point $(0,-1)$

Solving for the slope, where $y' =m$

$\begin{equation} \begin{aligned} y' = m &= \frac{d}{dx} \left( \frac{x^2-1}{x^2+1} \right) \\ \\ m &= \frac{(x^2+1)\frac{d}{dx}(x^2-1)-(x^2-1)\frac{d}{dx}(x^2+1) }{(x^2+1)^2}\\ \\ m &= \frac{(x^2+1)(2x)-(x^2-1)(2x)}{(x^2+1)^2}\\ \\ m &= \frac{\cancel{2x^3} + 2x - \cancel{2x^3} + 2x }{(x^2+1)}\\ \\ m &= \frac{4x}{(x^2+1)^2}\\ \\ m &= \frac{4(0)}{(0^2+1)^2}\\ \\ m &= \frac{0}{1}\\ \\ m &= 0 \end{aligned} \end{equation}$


Using point slope form


$\begin{equation} \begin{aligned} y - y_1 &= m(x-x_1)\\ \\ y - (-1) &= 0(x-0)\\ \\ y + 1 &= 0 \\ \\ y &= -1 && \Longleftarrow \text{(Equation of the tangent line to the curve at (0,-1))} \end{aligned} \end{equation}$