What is the tangent of f(x)= 2/(1+e^{-x)},(0,1) ?

The tangent of f(x)= 2/(1+e^{-x)},(0,1) is y= 1/2 x+1

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tangent of f(x)= 2/(1+e^{-x)},(0,1)

tangent of

f (x) = \frac{2}{1 + e ^{- x}}, (0, 1)

y = \frac{1}{2} x + 1

Solution steps

Find the slope of

f (x) = \frac{2}{1 + e ^{- x}} : \frac{df ( x )}{d x} = \frac{2 e ^{- x}}{( 1 + e ^{- x} ) ^{2}}

Compute the slope of

f (x) = \frac{2}{1 + e ^{- x}}

(0, 1) : m = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

(0, 1) : y = \frac{1}{2} x + 1

y = \frac{1}{2} x + 1

x^{^{'}} = (x - 17)^{2}

f (x) = \frac{7}{x + 8}, (5, \frac{7}{13})

y^{^{'}} = 0.06 (y - 65)

n = 1 \sum \infty \frac{n}{2 n + 1}

\frac{d}{d x} (e^{x} + e^{- 3 x})

What is the tangent of f(x)= 2/(1+e^{-x)},(0,1) ?
The tangent of f(x)= 2/(1+e^{-x)},(0,1) is y= 1/2 x+1