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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

Simplify

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

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

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

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

\frac{\partial}{\partial y} (e^{x} - 1)

\int (1 - sin^{2} (θ)) d θ

\frac{d}{d x} ((2 x^{2} - 1)^{3} (x^{4} + 3)^{5})

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