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

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

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

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

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Popular >

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

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

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

\int \frac{x - 2}{x + 1} d x

h \to 0 lim (\frac{f ( h ) ( 8 + h ) - f ( 8 )}{h})

(- 5, - 10), (- 1, 5)

\frac{( 2 - 5 s )}{s ^{2} + 9}

\frac{d}{d x} (\frac{1}{g ( x )})

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