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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\frac{1}{e ^{- x} + 1}

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d ^{2}}{d x ^{2}} (e^{8 x^{2}})

n = 1 \sum \infty 4^{- 2 n + 1}

\frac{d}{d x} (\frac{1}{4 - x ^{2}})

x \to 1 lim ((1 - x^{2})^{1 - x})

\frac{d}{d v} (u + v)

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