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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Popular >

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

Simplify

\frac{1}{\frac{e ^{x} - 1}{e ^{x} + 1}} \cdot \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 )}

x \to 7 lim (\frac{1}{x - 7})

\int_{- \infty}^{- 1} x^{- \frac{14}{5}} d x

\int \frac{x + 3}{x ^{2} + 6 x + 1} d x

\int y (2 y^{2} - 3) d y

\frac{d}{d x} (4^{- 2 x})

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))