What is the derivative of (e^x+1/(e^x-1)) ?

The derivative of (e^x+1/(e^x-1)) is -(2e^x)/((e^x-1)^2)

What is the first derivative of (e^x+1/(e^x-1)) ?

The first derivative of (e^x+1/(e^x-1)) is -(2e^x)/((e^x-1)^2)

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

f (x) = 9 x^{8} e^{x} + e^{x} x^{9}

\frac{d}{d x} (\frac{5 + csc ( x )}{9 - csc ( x )})

y^{^{'}} = x^{4} - \frac{1}{x} y

\int_{0}^{1} x^{2} ln (x) d x

\frac{d}{d x} (\frac{1}{( 1 - x ^{6} ) ^{\frac{3}{2}}})

What is the derivative of (e^x+1/(e^x-1)) ?
The derivative of (e^x+1/(e^x-1)) is -(2e^x)/((e^x-1)^2)
What is the first derivative of (e^x+1/(e^x-1)) ?
The first derivative of (e^x+1/(e^x-1)) is -(2e^x)/((e^x-1)^2)