What is the tangent of f(x)= 6/((1+e^{-x))},\at x=0 ?

The tangent of f(x)= 6/((1+e^{-x))},\at x=0 is y= 3/2 x+3

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tangent of f(x)= 6/((1+e^{-x))},\at x=0

tangent of

f (x) = \frac{6}{( 1 + e ^{- x} )}, at x = 0

y = \frac{3}{2} x + 3

Solution steps

Find the tangent point:

(0, 3)

Find the slope of

f (x) = \frac{6}{1 + e ^{- x}} : \frac{df ( x )}{d x} = \frac{6 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{3}{2}

Find the line with slope m=

\frac{3}{2}

and passing through

(0, 3) : y = \frac{3}{2} x + 3

y = \frac{3}{2} x + 3

\frac{d y}{d x} = \frac{y}{x} + 4 x + 1

\frac{\partial}{\partial x} (t e^{x t})

y = e^{5 e^{x}}

\int 4 x - 9 d x

h^{^{'}} (s) = 3 + 5 sin (s)

What is the tangent of f(x)= 6/((1+e^{-x))},\at x=0 ?
The tangent of f(x)= 6/((1+e^{-x))},\at x=0 is y= 3/2 x+3