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

Open Ningshiqi opened this issue 6 years ago • 4 comments

Ningshiqi avatar Sep 14 '18 02:09 Ningshiqi

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

hdsong1996 avatar Mar 26 '20 17:03 hdsong1996

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

try1995 avatar Dec 27 '20 02:12 try1995

tanh x is not the same as tan x

laluarif93 avatar Mar 07 '21 14:03 laluarif93

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

nicodjimenez avatar Mar 08 '21 04:03 nicodjimenez