What is the derivative of f(x)= 2/(1+e^{-x)} ?

The derivative of f(x)= 2/(1+e^{-x)} is (2e^{-x})/((1+e^{-x))^2}

What is the first derivative of f(x)= 2/(1+e^{-x)} ?

The first derivative of f(x)= 2/(1+e^{-x)} is (2e^{-x})/((1+e^{-x))^2}

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derivative of

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int 7 sin^{5} (x) cos^{- \frac{3}{2}} (x) d x

\int_{0}^{2} \int_{0}^{3} (\frac{x ^{2} y ^{2}}{4}) d x d y

f (x) = sin (5 ln (x))

\frac{d}{d x} (\frac{sin ( 5 x )}{x})

((cos (x))^{2})^{^{'}}

What is the derivative of f(x)= 2/(1+e^{-x)} ?
The derivative of f(x)= 2/(1+e^{-x)} is (2e^{-x})/((1+e^{-x))^2}
What is the first derivative of f(x)= 2/(1+e^{-x)} ?
The first derivative of f(x)= 2/(1+e^{-x)} is (2e^{-x})/((1+e^{-x))^2}