What is the derivative of 2ln|x-e^x| ?

The derivative of 2ln|x-e^x| is (2(-e^x+1))/(x-e^x)

What is the first derivative of 2ln|x-e^x| ?

The first derivative of 2ln|x-e^x| is (2(-e^x+1))/(x-e^x)

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\frac{d}{d x} (2 ln ∣ x - e^{x} ∣)

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

Solution steps

\frac{d}{d x} (2 ln ∣ x - e^{x} ∣)

Take the constant out:

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

= 2 \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} ∣}

= 2 \cdot \frac{1}{∣ x - e ^{x} ∣} \cdot \frac{( x - e ^{x} ) ( 1 - e ^{x} )}{∣ x - e ^{x} ∣}

Simplify

2 \cdot \frac{1}{∣ x - e ^{x} ∣} \cdot \frac{( x - e ^{x} ) ( 1 - e ^{x} )}{∣ x - e ^{x} ∣} : \frac{2 ( - e ^{x} + 1 )}{x - e ^{x}}

= \frac{2 ( - e ^{x} + 1 )}{x - e ^{x}}

\frac{\partial}{\partial y} (- \frac{1}{ln ( 10 ) ( x - y )})

f (x) = \frac{8}{sin ( x ) + cos ( x )}

f (x) = x^{2} - 12, g (x) = x - 6, [23, 6]

\int_{0}^{1} x - x d x

f (x) = 1 + cos (x), at x = \frac{3 π}{2}

What is the derivative of 2ln|x-e^x| ?
The derivative of 2ln|x-e^x| is (2(-e^x+1))/(x-e^x)
What is the first derivative of 2ln|x-e^x| ?
The first derivative of 2ln|x-e^x| is (2(-e^x+1))/(x-e^x)