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

The derivative of (e^x-e^{-x})/(e^x+e^{-x)} is 4/((e^x+e^{-x))^2}

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

The first derivative of (e^x-e^{-x})/(e^x+e^{-x)} is 4/((e^x+e^{-x))^2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

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

\frac{4}{( e ^{x} + e ^{- x} ) ^{2}}

Solution steps

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

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

Simplify

\frac{( e ^{x} + e ^{- x} ) ( e ^{x} + e ^{- x} ) - ( e ^{x} - e ^{- x} ) ( e ^{x} - e ^{- x} )}{( e ^{x} + e ^{- x} ) ^{2}} : \frac{4}{( e ^{x} + e ^{- x} ) ^{2}}

= \frac{4}{( e ^{x} + e ^{- x} ) ^{2}}

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

\frac{d}{d v} (v^{2})

\int 8 x^{3} (1 + x^{4})^{10} d x

\int \frac{1}{( x ^{2} x ^{2} - 16 )} d x

f (x) = 9 x^{\frac{1}{2}} + x^{\frac{3}{2}}, at x = 9

What is the derivative of (e^x-e^{-x})/(e^x+e^{-x)} ?
The derivative of (e^x-e^{-x})/(e^x+e^{-x)} is 4/((e^x+e^{-x))^2}
What is the first derivative of (e^x-e^{-x})/(e^x+e^{-x)} ?
The first derivative of (e^x-e^{-x})/(e^x+e^{-x)} is 4/((e^x+e^{-x))^2}