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

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

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

tangent of

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

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

Solution steps

Find the tangent point:

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

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{e}{( 1 + e ) ^{2}}

Find the line with slope m=

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

and passing through

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

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

\frac{d y}{d x} = 7 x e^{y}

d er i v a t i v e f (x) = 2 x^{6} (x^{4} - 2 x)

\int_{2}^{\infty} \frac{1}{5 x ^{2}} d x

t a y l or \frac{1}{1 - x}, 4

x \to 0 lim (6 x^{2})

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