Derivative of the Absolute Value

Definition

The derivative of the absolute value function |x| is the sign function: $$ D[ \: |x| \: ] = \frac{x}{|x|} = \frac{|x|}{x} $$

You can write the sign function either with |x| in the numerator or in the denominator.

Both forms ultimately give the same result.

Proof

The absolute value function

$$ f(x)=|x| $$

can also be written in the equivalent form

$$ f(x)=\sqrt{x^2} $$

Note. The absolute value of a variable |x| is equal to the square root of x squared.

Since the derivative of a square root is

$$ D[\sqrt{x}]=\frac{1}{2 \sqrt{x}} $$

we can apply this rule to f(x):

$$ D[\sqrt{x^2}]=\frac{2x}{2 \sqrt{x^2}}= \frac{x}{\sqrt{x^2}}$$

This yields the derivative of f(x):

$$ f'(x) = \frac{x}{\sqrt{x^2}}$$

Recognizing that the square root of $ x^2 $ is simply |x|, we can rewrite f'(x) as:

$$ f'(x) = \frac{x}{|x|}$$

Thus, we arrive at the derivative of |x|, which is the sign function, also denoted as sgn(x).

graph showing the derivative of the absolute value function

The sign function takes the value 1 when $ x $ is positive (x > 0) and -1 when $ x $ is negative (x < 0).

Note. The sign function is not differentiable at x = 0.

Alternative Method

There’s also an alternative way to prove the derivative of the absolute value function.

We can rewrite $ f(x) = |x| $ as a piecewise function:

$$ f(x) = \begin{cases} f(x)=x \:\: \text{if} \:\: x \gt 0 \\ f(x)=-x \:\: \text{if} \:\: x \lt 0 \end{cases} $$

Differentiating each piece separately, we get:

$$ f'(x) = \begin{cases} f'(x)=1 \:\: \text{if} \:\: x \gt 0 \\ f'(x)=-1 \:\: \text{if} \:\: x \lt 0 \end{cases} $$

This leads us to the same result: the derivative is the sign function:

$$ f'(x) = sgn \: x $$

And so on.