What is the derivative of 8/(e^x+e^{-x}) ?

The derivative of 8/(e^x+e^{-x}) is -(8(e^x-e^{-x}))/((e^x+e^{-x))^2}

What is the first derivative of 8/(e^x+e^{-x}) ?

The first derivative of 8/(e^x+e^{-x}) is -(8(e^x-e^{-x}))/((e^x+e^{-x))^2}

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\frac{d}{d x} (\frac{8}{e ^{x} + e ^{- x}})

- \frac{8 ( e ^{x} - e ^{- x} )}{( e ^{x} + e ^{- x} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{8}{e ^{x} + e ^{- x}})

Take the constant out:

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

= 8 \frac{d}{d x} (\frac{1}{e ^{x} + e ^{- x}})

Apply exponent rule:

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

= 8 \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}

= 8 (- \frac{1}{( e ^{x} + e ^{- x} ) ^{2}} (e^{x} - e^{- x}))

Simplify

8 (- \frac{1}{( e ^{x} + e ^{- x} ) ^{2}} (e^{x} - e^{- x})) : - \frac{8 ( e ^{x} - e ^{- x} )}{( e ^{x} + e ^{- x} ) ^{2}}

= - \frac{8 ( e ^{x} - e ^{- x} )}{( e ^{x} + e ^{- x} ) ^{2}}

\int \frac{e ^{3 x}}{4 - 4 e ^{6 x}} d x

\frac{d}{d z} ((2 z + 3 i) (z - i))

x \to 0 lim (\frac{3 x}{sin ( x )})

\int (x + 1) (x - x + 1) d x

\int t cos (t^{2}) d t

What is the derivative of 8/(e^x+e^{-x}) ?
The derivative of 8/(e^x+e^{-x}) is -(8(e^x-e^{-x}))/((e^x+e^{-x))^2}
What is the first derivative of 8/(e^x+e^{-x}) ?
The first derivative of 8/(e^x+e^{-x}) is -(8(e^x-e^{-x}))/((e^x+e^{-x))^2}