The derivative is: \(1-\tanh ^{2}(x)\)
Hyperbolic functions work in the same way as the "normal" trigonometric "cousins" but instead of referring to a unit circle (for \(\sin , \cos\) and \(\tan\) ) they refer to a set of hyperbolae.
You can write:
\[
\tanh (x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}
\]
It is now possible to derive using the rule of the quotient and the fact that: derivative of \(e^{x}\) is \(e^{x}\) and derivative of \(e^{-x}\) is \(-e^{-x}\)
So you have:
\[
\begin{aligned}
&\frac{d}{d x} \tanh (x)=\frac{\left(e^{x}+e^{-x}\right)\left(e^{x}+e^{-x}\right)-\left(e^{x}-e^{-x}\right)\left(e^{x}-e^{-x}\right)}{\left(e^{x}+e^{-x}\right)^{2}} \\
&=1-\frac{\left(e^{x}-e^{-x}\right)^{2}}{\left(e^{x}+e^{-x}\right)^{2}}=1-\tanh ^{2}(x)
\end{aligned}
\]