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

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

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

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

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

\frac{e ^{x}}{e ^{x} - 1}

Solution steps

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

Apply the chain rule:

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

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

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

= \frac{1}{e ^{x} - 1} e^{x}

Simplify

\frac{1}{e ^{x} - 1} e^{x} : \frac{e ^{x}}{e ^{x} - 1}

= \frac{e ^{x}}{e ^{x} - 1}

f (x) = \frac{1}{x}, at x = a

\frac{d}{d p} (\frac{p - 1}{p ^{2} + 4})

\int \frac{e ^{x}}{1 + e ^{x} + e ^{2 x}} d x

\int 6 x - 7 d x

\frac{\partial}{\partial x} (- e^{- o} \cdot cos (o))

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