What is (\partial)/(\partial x)(-ln(1+e^{-x})) ?

The answer to (\partial)/(\partial x)(-ln(1+e^{-x})) is (e^{-x})/(1+e^{-x)}

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(\partial)/(\partial x)(-ln(1+e^{-x}))

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

\frac{e ^{- x}}{1 + e ^{- x}}

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

= \frac{e ^{- x}}{1 + e ^{- x}}

f (w) = (6 x - 8)^{2}

f (x) = 7 arctan (x - 1 + x^{2})

f (x) = x^{x^{x}}

y = 5 x^{\frac{1}{2}} + x^{\frac{3}{2}}, at x = 25

\frac{\partial}{\partial x} (ln (3))

What is (\partial)/(\partial x)(-ln(1+e^{-x})) ?
The answer to (\partial)/(\partial x)(-ln(1+e^{-x})) is (e^{-x})/(1+e^{-x)}