What is the derivative of 2/(1+e^{-x}) ?

The derivative of 2/(1+e^{-x}) is (2e^{-x})/((1+e^{-x))^2}

What is the first derivative of 2/(1+e^{-x}) ?

The first derivative of 2/(1+e^{-x}) is (2e^{-x})/((1+e^{-x))^2}

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

n = 0 \sum \infty 3 (- \frac{1}{2})^{n}

x \to 2 lim (\frac{4}{( x - 2 ) ^{5}})

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

\int - \frac{2 x}{x ^{4} - 1} d x

x \to \frac{π}{3} lim (cos (x))

What is the derivative of 2/(1+e^{-x}) ?
The derivative of 2/(1+e^{-x}) is (2e^{-x})/((1+e^{-x))^2}
What is the first derivative of 2/(1+e^{-x}) ?
The first derivative of 2/(1+e^{-x}) is (2e^{-x})/((1+e^{-x))^2}