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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

= 2 \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}

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

Simplify

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

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

y = 4 x, y = 3 x^{2}

x \frac{d y}{d x} - 3 y = 6

\int_{- \frac{π}{6}}^{0} 8 sec^{3} (x) d x

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

\int e^{5 x} cos (6 x) d x

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