Posted by dino on Monday, November 5, 2012 at 8:51pm.
y = tanh(x)
there are asymptotes at y=1 and y=-1
just consider when x gets large positive. e^-x vanishes, and you have e^x/e^x = 1
when x gets large negative, e^x vanishes, and you have -e^-x/e^-1 = -1
first of all simplify it a bit,
multiplying top and bottom by e^x to get
y = (e^(2x) -1)/(e^(2x) + 1)
dy/dx = [(e^(2x) + 1)(2e^(2x)) - (e^(2x) -1)(2e^(2x)) )/(e^(2x) + 1)^2
= 2e^(2x)(e^(2x) + 1 + e^(2x) - 1)/(e^(2x+1)^2
= 4e^(2x)/(e^(2x)+1)^2
confirmed by Wolfram:
http://www.wolframalpha.com/input/?i=derivative+of+%28e^x+-+e^-x%29%2F%28e^x+%2B+e^-x%29
I will leave it up to you to sub in x=1 and x=-1 to show that you get the same value, I did.
thank you very Much
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