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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derivative of

\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}}

e^{- 6 x}, 0

\int x ln (5 x) d x

x \to 2 - lim (6 - 3 x - 3)

\int_{0}^{1} 2 x (8 - x^{2}) d x

x \to - 3 lim (- 3 x)

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}