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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derivative of

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 )}

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))