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What is the derivative of tanh(x)?
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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}
by Diamond (38,983 points)

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