What is the derivative of 5/(e^x+e^{-x)} ?

The derivative of 5/(e^x+e^{-x)} is -(5(e^x-e^{-x}))/((e^x+e^{-x))^2}

What is the first derivative of 5/(e^x+e^{-x)} ?

The first derivative of 5/(e^x+e^{-x)} is -(5(e^x-e^{-x}))/((e^x+e^{-x))^2}

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derivative of

\frac{5}{e ^{x} + e ^{- x}}

- \frac{5 ( e ^{x} - e ^{- x} )}{( e ^{x} + e ^{- x} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{5}{e ^{x} + e ^{- x}})

Take the constant out:

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

= 5 \frac{d}{d x} (\frac{1}{e ^{x} + e ^{- x}})

Apply exponent rule:

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

= 5 \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}

= 5 (- \frac{1}{( e ^{x} + e ^{- x} ) ^{2}} (e^{x} - e^{- x}))

Simplify

5 (- \frac{1}{( e ^{x} + e ^{- x} ) ^{2}} (e^{x} - e^{- x})) : - \frac{5 ( e ^{x} - e ^{- x} )}{( e ^{x} + e ^{- x} ) ^{2}}

= - \frac{5 ( e ^{x} - e ^{- x} )}{( e ^{x} + e ^{- x} ) ^{2}}

2 x^{2} + 2 x + x y = 2

f (x) = x^{- \frac{1}{9}}

\frac{( s - 5 )}{( s ^{2} - 6 s + 25 )}

\frac{d}{d x} (\frac{6 x ^{7}}{5 tan ( x )})

\frac{d}{d x} (\frac{2 x + 5 x ^{- \frac{2}{5}}}{x ^{2} + 1})

What is the derivative of 5/(e^x+e^{-x)} ?
The derivative of 5/(e^x+e^{-x)} is -(5(e^x-e^{-x}))/((e^x+e^{-x))^2}
What is the first derivative of 5/(e^x+e^{-x)} ?
The first derivative of 5/(e^x+e^{-x)} is -(5(e^x-e^{-x}))/((e^x+e^{-x))^2}