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September 23, 2014

September 23, 2014

Posted by **dino** on Monday, November 5, 2012 at 8:51pm.

I know to use the Quotient rule but I am having a hard time with this problem.

- calculus -
**Steve**, Monday, November 5, 2012 at 9:17pmy = 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

- calculus -
**Reiny**, Monday, November 5, 2012 at 9:30pmfirst 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.

- calculus -
**dino**, Monday, November 5, 2012 at 9:35pmthank you very Much

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