What is the derivative of ln|x-1| ?

The derivative of ln|x-1| is 1/(x-1)

What is the first derivative of ln|x-1| ?

The first derivative of ln|x-1| is 1/(x-1)

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\frac{d}{d x} (ln ∣ x - 1 ∣)

\frac{1}{x - 1}

Solution steps

\frac{d}{d x} (ln ∣ x - 1 ∣)

Apply the chain rule:

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

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

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

= \frac{1}{∣ x - 1 ∣} \cdot \frac{x - 1}{∣ x - 1 ∣}

Simplify

\frac{1}{∣ x - 1 ∣} \cdot \frac{x - 1}{∣ x - 1 ∣} : \frac{1}{x - 1}

= \frac{1}{x - 1}

f^{^{'}} (x) = cos (x), f (0) = - 5

\frac{d}{d x} (sin^{2} (- 3 x + 6))

(- 10, 3), (- 8, - 8)

n = 0 \sum \infty (\frac{1}{5})^{n + 1}

\int (e^{x} + 2 x)^{2} d x

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