calculus

Show that the tanget lines to the graph of Y= e^x - e^-x /e^x + e^-x at x =1 and x= -1 are parallel.
I know to use the Quotient rule but I am having a hard time with this problem.

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asked by dino
  1. 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

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    posted by Steve
  2. 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.

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    posted by Reiny
  3. thank you very Much

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    posted by dino

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