What is the tangent of f(x)= 1/(1+e^x),\at x=1 ?

The tangent of f(x)= 1/(1+e^x),\at x=1 is y=-e/((1+e)^2)x+(2e+1)/((e+1)^2)

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tangent of f(x)= 1/(1+e^x),\at x=1

tangent of

f (x) = \frac{1}{1 + e ^{x}}, at x = 1

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

Solution steps

Find the tangent point:

(1, \frac{1}{1 + e})

Find the slope of

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

Compute the slope of

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

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

Find the line with slope m=

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

and passing through

(1, \frac{1}{1 + e}) : y = - \frac{e}{( 1 + e ) ^{2}} x + \frac{2 e + 1}{( e + 1 ) ^{2}}

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

\int \frac{6 - 2 x}{8 - 4 x - 4 x ^{2}} d x

\frac{d}{d z} (sin (y z))

\int cot (\frac{x}{2}) d x

f (- 2) = 3 x - 7

f (x) = sin (π x)

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