What is the tangent of y=(10)/(1+e^{-x)},(0,5) ?

The tangent of y=(10)/(1+e^{-x)},(0,5) is y= 5/2 x+5

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tangent of y=(10)/(1+e^{-x)},(0,5)

tangent of

y = \frac{10}{1 + e ^{- x}}, (0, 5)

y = \frac{5}{2} x + 5

Solution steps

Find the slope of

y = \frac{10}{1 + e ^{- x}} : \frac{d y}{d x} = \frac{10 e ^{- x}}{( 1 + e ^{- x} ) ^{2}}

Compute the slope of

y = \frac{10}{1 + e ^{- x}}

(0, 5) : m = \frac{5}{2}

Find the line with slope m=

\frac{5}{2}

and passing through

(0, 5) : y = \frac{5}{2} x + 5

y = \frac{5}{2} x + 5

(3 x^{3} + 3) (x^{4} - 3 x)

\int_{0}^{5} x e^{- 4 x} d x

\int \frac{2}{9 - 4 x ^{2}} d x

\int_{1}^{3} \frac{1}{( x - 3 ) ^{2}} d x

\int 7 v (v^{2} + 4)^{2} d v

What is the tangent of y=(10)/(1+e^{-x)},(0,5) ?
The tangent of y=(10)/(1+e^{-x)},(0,5) is y= 5/2 x+5