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}
$