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)

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

- \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}}

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