## Calculus: Early Transcendentals (2nd Edition)

Slope of Tangent Line: $\frac{1}{5}$
$f(x) = \tan^{-1}(x)$ $f'(x) = \frac{1}{1+x^2}$ $f'(-2)=\frac{1}{1+(-2)^2} = \frac{1}{5}$