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}

\frac{d}{d x} (x^{2} - 2 x)

\frac{d}{d x} (\frac{x - 4}{3 x - x ^{2}})

\int 2 + 2 cos (x) d x

\int \frac{u ^{2}}{1 + u ^{2}} d u

y = 2 x^{2}, y = 8 x^{2}, 5 x + y = 3

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)