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

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

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

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

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derivative of

y = \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}}

y = x^{5 s i n (x)}

y^{^{'}} = \frac{( x ^{3} y ^{3} + 2 x ^{3} )}{( y ^{2} )}

\frac{2}{( s - 2 ) ^{4}}

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

f (x) = sin^{2} (x) cos (x)

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