What is the derivative of 4ln|x-e^x| ?

The derivative of 4ln|x-e^x| is (4(-e^x+1))/(x-e^x)

What is the first derivative of 4ln|x-e^x| ?

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

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\frac{d}{d x} (4 ln ∣ x - e^{x} ∣)

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

Solution steps

\frac{d}{d x} (4 ln ∣ x - e^{x} ∣)

Take the constant out:

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

= 4 \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} ∣}

= 4 \cdot \frac{1}{∣ x - e ^{x} ∣} \cdot \frac{( x - e ^{x} ) ( 1 - e ^{x} )}{∣ x - e ^{x} ∣}

Simplify

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

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

y = \frac{cos ( x )}{1 - sin ( x )}

y = cos (sin (tan (9 x)))

\frac{\partial}{\partial x} (4 - x^{2} + x y - 2 y^{2})

x^{2} + y^{2} = 29, (- 5, 2)

\int_{0}^{1} \frac{x ^{2}}{1 + x ^{6}} d x

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