Question

What is the derivative of `tanh(x)`?

Answer 1

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.

(Picture source: Physicsforums.com)

You can write:
`tanh(x)=(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:
`d/dxtanh(x)=[(e^x+e^-x)(e^x+e^-x)-(e^x-e^-x)(e^x-e^-x)]/(e^x+e^-x)^2`
`=1-((e^x-e^-x)^2)/(e^x+e^-x)^2=1-tanh^2(x)`