Derivative of ln of the absolute value of x

Statement

Verify that $D[\ln|x|] = 1/x$ both for $x>0$ and for $x<0$.

Solution

The two cases are distinguished according to the sign of $x$.

Case $x>0$: $|x|=x$, so $\ln|x|=\ln x$ and $D[\ln x] = \dfrac{1}{x}$.

Case $x<0$: $|x|=-x$, so $\ln|x|=\ln(-x)$. By the chain rule: $D[\ln(-x)] = \frac{1}{-x}\cdot(-1) = \frac{1}{x}.$ In both cases $D[\ln|x|] = \boxed{\dfrac{1}{x}}$ (for every $x\ne 0$).

Links

Topics: Derivatives
Concepts: Composite derivative · Absolute value
Skills: Differentiating · Reasoning by cases
Exercise type: Derivative computation