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}}

sin (cos (7 x))

x \to 0 lim (\frac{x ^{2} - x + 1}{x})

y^{^{'}} + 6 y = 2 t + 3

\frac{s + 4}{( s + 2 ) ^{2}}

\frac{d y}{d x} = - \frac{x}{y e ^{x^{2}}}

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)