So I figured I can use the chain rule to do this:
$g\prime(x)=\frac{1}{f^\prime(g(x))}$
So that
$(\arctan(x))\prime = \frac{1}{\left[\sec^2(\arctan(x)){}\right]^\prime}$
But this book tells me that
$(\arctan(x))\prime = \frac{1}{x^2+1}$
So, how do I show that $1+x^2=\sec^2(\arctan(x))$?