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The tanh_derivative should be : 1. + values**2 #41

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Ningshiqi opened this issue Sep 14, 2018 · 4 comments

Ningshiqi commented Sep 14, 2018

No description provided.

hdsong1996 commented Mar 26, 2020

No， he is right.[tanh(x)]'=1-[tanh(x)]^2

try1995 commented Dec 27, 2020

[tanh(x)]' = [sin(x)/cos(x)]' = (cos(x)^2 + sin(x)^2)/cos(x)^2 = 1 + tanx(x)^2

laluarif93 commented Mar 7, 2021

tanh x is not the same as tan x

Owner

nicodjimenez commented Mar 8, 2021

It's been a while since I've looked at this codebase... if there's something wrong can someone submit a PR?

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