What is the derivative of tanh(x)?

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1 Answer
Gió
Dec 22, 2014

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.

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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)

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