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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