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!

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