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

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

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

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

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Popular >

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

y^{^{'}} = e^{y} 5^{x}

n = 1 \sum \infty \frac{60}{6 ^{n}}

\int e^{- 0.08 x} d x

\frac{d}{d x} (tan^{2} (x) cos (x))

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