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
To start, we need to find the equation of the tangent line at x = 1 and x = -1.
Let's begin by finding the derivative of the function Y = (e^x - e^(-x))/(e^x + e^(-x)).
First, simplify the function:
Y = (e^x - e^(-x))/(e^x + e^(-x))
= (e^x - e^(-x))^2 / (e^x + e^(-x)) * (e^x - e^(-x))/(e^x - e^(-x))
= (e^(2x) - 2 + e^(-2x))/(e^x - e^(-x))
Now, differentiate Y using the Quotient Rule:
dY/dx = [(2e^(2x) + 2e^(-2x))(e^x - e^(-x)) - (e^(2x) - 2 + e^(-2x))(e^x + e^(-x))]/((e^x - e^(-x))^2)
Simplify the expression:
dY/dx = [(2e^(2x) + 2e^(-2x))(e^x - e^(-x)) - (e^(2x) - 2 + e^(-2x))(e^x + e^(-x))] / (e^(2x) - 2 + e^(-2x))
Now, substitute x = 1 and x = -1 into the derivative to find the slopes of the tangent lines:
Slope of tangent line at x = 1:
m1 = [(2e^2 + 2e^(-2))(e - e^(-1)) - (e^2 - 2 + e^(-2))(e + e^(-1))] / (e^(2) - 2 + e^(-2))
Slope of tangent line at x = -1:
m2 = [(2e^(-2) + 2e^(2))(e^(-1) - e) - (e^(-2) - 2 + e^(2))(e^(-1) + e)] / (e^(-2) - 2 + e^(2))
To show that the tangent lines are parallel, we need to show that m1 = m2.
You can now substitute the values of e (approximately 2.71828) into the expressions for m1 and m2 to calculate their values. If m1 and m2 are equal, then the tangent lines will be parallel.