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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d θ} (ln (x (θ)) (- θ - 2))

\int - csc^{2} (\frac{9 x}{8}) d x

\int e^{u} d u

9 \cdot x^{2} \cdot y^{^{'}} = y^{^{'}} + 7 \cdot x \cdot e^{- y}

f (x) = 6 x^{\frac{4}{3}} - \frac{2}{3} x^{\frac{3}{2}} + x^{2} - 7 x + 1

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