calculus
posted by dino on .
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.

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