How do you find the equation of the tangent line to the curve #y=2/(1+e^-x)# at (0,1)?

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Answer 1 · 2017-02-02T17:58:08

#y = 1 - x#

Differentiate using the quotient rule.

#y' = (0 * (1 + e^-x) - (-2e^(-x)))/(1 + e^(-x))^2#

#y' = (2e^(-x))/(1 + e^(-x))^2#

Find the slope of the tangent by plugging in your point #x= a# into the derivative.

#y' = 2/(1 + e^0)#

#y' = 2/2#

#y' = 1#

Now find the equation.

#y - y_1 = m(x -x_1)#

#y - 1 = 1(x - 0)#

#y - 1 = -x#

#y = 1 - x#

Hopefully this helps!