What is the derivative of ln(x+4+e^{-3x}) ?

The derivative of ln(x+4+e^{-3x}) is (-3e^{-3x}+1)/(x+4+e^{-3x)}

What is the first derivative of ln(x+4+e^{-3x}) ?

The first derivative of ln(x+4+e^{-3x}) is (-3e^{-3x}+1)/(x+4+e^{-3x)}

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derivative of

ln (x + 4 + e^{- 3 x})

\frac{- 3 e ^{- 3 x} + 1}{x + 4 + e ^{- 3 x}}

Solution steps

\frac{d}{d x} (ln (x + 4 + e^{- 3 x}))

Apply the chain rule:

\frac{1}{x + 4 + e ^{- 3 x}} \frac{d}{d x} (x + 4 + e^{- 3 x})

= \frac{1}{x + 4 + e ^{- 3 x}} \frac{d}{d x} (x + 4 + e^{- 3 x})

\frac{d}{d x} (x + 4 + e^{- 3 x}) = - 3 e^{- 3 x} + 1

= \frac{1}{x + 4 + e ^{- 3 x}} (- 3 e^{- 3 x} + 1)

Simplify

\frac{1}{x + 4 + e ^{- 3 x}} (- 3 e^{- 3 x} + 1) : \frac{- 3 e ^{- 3 x} + 1}{x + 4 + e ^{- 3 x}}

= \frac{- 3 e ^{- 3 x} + 1}{x + 4 + e ^{- 3 x}}

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

n = 0 \sum \infty \frac{x ^{n}}{n ^{n}}

x \to 0 lim (\frac{4 ^{x} - 3 ^{x}}{x})

\int \frac{1}{3} sin (3 x) d x

What is the derivative of ln(x+4+e^{-3x}) ?
The derivative of ln(x+4+e^{-3x}) is (-3e^{-3x}+1)/(x+4+e^{-3x)}
What is the first derivative of ln(x+4+e^{-3x}) ?
The first derivative of ln(x+4+e^{-3x}) is (-3e^{-3x}+1)/(x+4+e^{-3x)}