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Popular Calculus >

(1/(1+e^{-x)})^'

Solution

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

Solution

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

Solution steps

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

Apply exponent rule:

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

= ((1 + e^{- x})^{- 1})^{^{'}}

Apply the chain rule:

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

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

(1 + e^{- x})^{'} = - e^{- x}

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

Simplify

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

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

Graph

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Frequently Asked Questions (FAQ)

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