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

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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{x + 1}{e ^{x}} d x

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

\int \frac{x}{- x - 1} d x

\int 4 x^{3} + 2 x y^{2} ln (x^{2} + y^{2}) d x

(x^{2} - x y) d x + x d y = 0

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