The Chain Rule implies that
\frac{d}{dx}(\arctan(\tanh(x)))=\frac{1}{1+\tanh^{2}(x)}\cdot \frac{d}{dx}(\tanh(x)).
Since \frac{d}{dx}(\tanh(x))=\sech^{2}(x), this becomes
\frac{d}{dx}(\arctan(\tanh(x)))=\frac{\sech^{2}(x)}{1+\tanh^{2}(x)}.
To see that this equals \sech(2x), you could note that
\sech(2x)=\frac{1}{\cosh(2x)}=\frac{2}{e^{2x}+e^{-2x}}
and
\frac{\sech^{2}(x)}{1+\tanh^{2}(x)}=\frac{\frac{1}{cosh^2(x)}}{1+\frac{\sinh^{2}(x)}{\cosh^{2}(x)}}=\frac{1}{\cosh^{2}(x)+\sinh^{2}(x)}
But this is the same as
\frac{1}{(\frac{e^{x}+e^{-x}}{2})^2+(\frac{e^{x}-e^{-x}}{2})^2}=\frac{4}{e^{2x}+2+e^{-2x}+e^{2x}-2+e^(-2x)}=\frac{2}{e^{2x}+e^{-2x}}.
