What is (\partial)/(\partial x)(1/(1+e^{-x)}) ?

The answer to (\partial)/(\partial x)(1/(1+e^{-x)}) is (e^{-x})/((1+e^{-x))^2}

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(\partial)/(\partial x)(1/(1+e^{-x)})

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{\partial}{\partial 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}}

\int \frac{sin ( 3 x )}{3} d x

\int \frac{- x}{1 + x} d x

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

\int_{0}^{2} x d x

f (x) = 1 0^{x}

What is (\partial)/(\partial x)(1/(1+e^{-x)}) ?
The answer to (\partial)/(\partial x)(1/(1+e^{-x)}) is (e^{-x})/((1+e^{-x))^2}