What is the derivative of 3ln|x-e^x| ?

The derivative of 3ln|x-e^x| is (3(-e^x+1))/(x-e^x)

What is the first derivative of 3ln|x-e^x| ?

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

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\frac{d}{d x} (3 ln ∣ x - e^{x} ∣)

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

Solution steps

\frac{d}{d x} (3 ln ∣ x - e^{x} ∣)

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 3 \frac{d}{d x} (ln ∣ x - e^{x} ∣)

Apply the chain rule:

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

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

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

= 3 \cdot \frac{1}{∣ x - e ^{x} ∣} \cdot \frac{( x - e ^{x} ) ( 1 - e ^{x} )}{∣ x - e ^{x} ∣}

Simplify

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

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

\int x^{2} \cdot (x - 5) + 7 - x d x

\int - 4 e^{x} d x

\int_{0}^{2} 2 π x (3 x (x - 2)^{2}) d x

n = 1 \sum \infty \frac{2 n}{3 n - 1}

\frac{d}{d x} (\frac{3 a}{4 x ^{5}})

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