Derivative of |x|: Formula, Proof and Examples

The derivative of \lvert x\rvert is \dfrac{x}{\lvert x\rvert}. In plain terms, the slope is 1 to the right of 0, -1 to the left of 0, and there is no slope at 0 itself.

Graph of |x| and its derivative

Proof, piece by piece

Split \lvert x\rvert into its two pieces. For x > 0, \lvert x\rvert = x, so the derivative is 1. For x < 0, \lvert x\rvert = -x, so the derivative is -1. In both cases the answer equals \frac{x}{\lvert x\rvert}: dividing a number by its absolute value just gives its sign.

Why there is no derivative at 0

Look at the difference quotient at x = 0:

From the right (h > 0) this is 1. From the left (h < 0) it is -1. The one-sided limits disagree, so the limit does not exist. The graph has a sharp corner, and no single tangent line fits it. This is the textbook example in the lesson on differentiability.

Another way to write the result is \frac{d}{dx}\lvert x\rvert = \frac{x}{\sqrt{x^2}}, since \lvert x\rvert = \sqrt{x^2}. Differentiating \sqrt{x^2} with the chain rule gives \frac{2x}{2\sqrt{x^2}}, the same formula.

The chain rule version: derivative of |u|

Example 1. \frac{d}{dx}\lvert 3x\rvert = \frac{3x}{\lvert 3x\rvert}\cdot 3, which is 3 for x > 0 and -3 for x < 0.

Example 2. \frac{d}{dx}\lvert x^2 - 4\rvert = \frac{(x^2 - 4)\cdot 2x}{\lvert x^2 - 4\rvert}. For \lvert x\rvert > 2 this is 2x. For \lvert x\rvert < 2 it is -2x. At x = \pm 2 the graph has corners and no derivative.

Example 3. \frac{d}{dx}\lvert\sin x\rvert = \frac{\sin x\cos x}{\lvert\sin x\rvert}, undefined wherever \sin x = 0.

A function built from |x| that is differentiable

Multiplying by x smooths the corner. Take f(x) = x\lvert x\rvert. For x > 0 it is x^2, for x < 0 it is -x^2, so f'(x) = 2\lvert x\rvert. At 0 the difference quotient is \frac{h\lvert h\rvert}{h} = \lvert h\rvert \to 0, so f'(0) = 0 exists. This f is also an antiderivative of 2\lvert x\rvert, which is how the integral of |x| is found.

Common mistakes

1. Answering \lvert 1\rvert = 1. The derivative is not 1 everywhere. On the left side it is -1.
2. Saying f'(0) = 0. The corner has no tangent line, so the derivative at 0 does not exist.
3. Thinking continuous means differentiable. \lvert x\rvert is continuous at 0 but still has no derivative there.

For the function itself, see the absolute value function page.