What is the derivative of 5ln|x-e^x| ?

The derivative of 5ln|x-e^x| is (5(-e^x+1))/(x-e^x)

What is the first derivative of 5ln|x-e^x| ?

The first derivative of 5ln|x-e^x| is (5(-e^x+1))/(x-e^x)

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\frac{d}{d x} (5 ln ∣ x - e^{x} ∣)

\frac{5 ( - e ^{x} + 1 )}{x - e ^{x}}

Solution steps

\frac{d}{d x} (5 ln ∣ x - e^{x} ∣)

Take the constant out:

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

= 5 \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} ∣}

= 5 \cdot \frac{1}{∣ x - e ^{x} ∣} \cdot \frac{( x - e ^{x} ) ( 1 - e ^{x} )}{∣ x - e ^{x} ∣}

Simplify

5 \cdot \frac{1}{∣ x - e ^{x} ∣} \cdot \frac{( x - e ^{x} ) ( 1 - e ^{x} )}{∣ x - e ^{x} ∣} : \frac{5 ( - e ^{x} + 1 )}{x - e ^{x}}

= \frac{5 ( - e ^{x} + 1 )}{x - e ^{x}}

z \to 1 lim (\frac{z}{2 i})

\frac{d}{d x} (- \frac{3 x + 1}{x ^{2}})

4 (5 - 7 x)^{5}

x \cdot sin (π x)

\int - 3 e^{x} d x

What is the derivative of 5ln|x-e^x| ?
The derivative of 5ln|x-e^x| is (5(-e^x+1))/(x-e^x)
What is the first derivative of 5ln|x-e^x| ?
The first derivative of 5ln|x-e^x| is (5(-e^x+1))/(x-e^x)