What is the derivative of f(x)=(14)/(1+e^{-x)} ?

The derivative of f(x)=(14)/(1+e^{-x)} is (14e^{-x})/((1+e^{-x))^2}

What is the first derivative of f(x)=(14)/(1+e^{-x)} ?

The first derivative of f(x)=(14)/(1+e^{-x)} is (14e^{-x})/((1+e^{-x))^2}

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derivative of

f (x) = \frac{14}{1 + e ^{- x}}

\frac{14 e ^{- x}}{( 1 + e ^{- x} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{14}{1 + e ^{- x}})

Take the constant out:

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

= 14 \frac{d}{d x} (\frac{1}{1 + e ^{- x}})

Apply exponent rule:

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

= 14 \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}

= 14 (- \frac{1}{( 1 + e ^{- x} ) ^{2}} (- e^{- x}))

Simplify

14 (- \frac{1}{( 1 + e ^{- x} ) ^{2}} (- e^{- x})) : \frac{14 e ^{- x}}{( 1 + e ^{- x} ) ^{2}}

= \frac{14 e ^{- x}}{( 1 + e ^{- x} ) ^{2}}

What is the derivative of f(x)=(14)/(1+e^{-x)} ?
The derivative of f(x)=(14)/(1+e^{-x)} is (14e^{-x})/((1+e^{-x))^2}
What is the first derivative of f(x)=(14)/(1+e^{-x)} ?
The first derivative of f(x)=(14)/(1+e^{-x)} is (14e^{-x})/((1+e^{-x))^2}