What is the derivative of 6/(1+e^{-x}) ?

The derivative of 6/(1+e^{-x}) is (6e^{-x})/((1+e^{-x))^2}

What is the first derivative of 6/(1+e^{-x}) ?

The first derivative of 6/(1+e^{-x}) is (6e^{-x})/((1+e^{-x))^2}

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\frac{d}{d x} (\frac{6}{1 + e ^{- x}})

\frac{6 e ^{- x}}{( 1 + e ^{- x} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6}{1 + e ^{- x}})

Take the constant out:

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

= 6 \frac{d}{d x} (\frac{1}{1 + e ^{- x}})

Apply exponent rule:

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

= 6 \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}

= 6 (- \frac{1}{( 1 + e ^{- x} ) ^{2}} (- e^{- x}))

Simplify

6 (- \frac{1}{( 1 + e ^{- x} ) ^{2}} (- e^{- x})) : \frac{6 e ^{- x}}{( 1 + e ^{- x} ) ^{2}}

= \frac{6 e ^{- x}}{( 1 + e ^{- x} ) ^{2}}

\int_{1}^{6} \frac{x ^{2} + 4}{7 x - x ^{2}} d x

\frac{\partial}{\partial x} (yz ln (x))

\frac{d}{d x} ((\frac{1}{3})^{x})

\int \frac{1}{e ^{x} - 2} d x

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

What is the derivative of 6/(1+e^{-x}) ?
The derivative of 6/(1+e^{-x}) is (6e^{-x})/((1+e^{-x))^2}
What is the first derivative of 6/(1+e^{-x}) ?
The first derivative of 6/(1+e^{-x}) is (6e^{-x})/((1+e^{-x))^2}