Absolute value rule · Derivative Calculator

What it says

$\left|x\right|$ is $x$ for positive $x$ and $-x$ for negative $x$, so its slope is $+1$ on the right and $-1$ on the left. That is exactly the sign function, $\operatorname{sgn} x$.

At $x = 0$ there is no derivative at all. Approaching from the right the difference quotient is $+1$, from the left $-1$; the one-sided limits disagree, so the two-sided limit does not exist. The graph has a corner, not a tangent.

For a compound argument the chain rule applies as usual: $\frac{d}{dx}\left|u\right| = \operatorname{sgn}(u)\,u'$, valid wherever $u \neq 0$.

When it applies

$\left|x\right|$ or $\texttt{abs(x)}$, alone or inside a larger expression.
Anywhere the argument is nonzero — the answer is undefined at the zeros of the argument.
Also covers $\sqrt{x^{2}}$, which is $\left|x\right|$ rather than $x$.

Five worked examples

Every line is the step the calculator would show, in the order it applies them. Each graph is live: hover it to read both curves and see the tangent whose slope is the derivative, drag to pan, scroll to zoom.

Example 1

\frac{d}{dx}\left[\left|x\right|\right]

Split by sign: the graph is $-x$ then $x$.
$\left|x\right| = \begin{cases} x & x > 0 \\ -x & x < 0\end{cases}$
Each piece is a line, with slope $+1$ or $-1$.
$\frac{d}{dx}\left|x\right| = \operatorname{sgn} x$

Answer

\operatorname{sgn}\left(x\right)

The V and its step. f′ jumps from $-1$ to $+1$ at the origin with no value in between, and that gap is the whole point: a corner has no tangent. Hover just either side and the readout flips sign without passing through anything.

Undefined at $x = 0$, where the two slopes meet at a corner.

Open abs(x) in the calculator

Example 2

\frac{d}{dx}\left[\left|x - 3\right|\right]

Chain rule with $u = x - 3$, and $u' = 1$.
$= \operatorname{sgn}\left(x - 3\right)\cdot 1$

Answer

\operatorname{sgn}\left(x - 3\right)

The same pair shifted to $x = 3$, where the step in f′ marks the corner in f.

The corner has moved to $x = 3$; the derivative jumps from $-1$ to $+1$ there.

Open abs(x-3) in the calculator

Example 3

\frac{d}{dx}\left[x \left|x\right|\right]

Product rule.
$= 1\cdot\left|x\right| + x\operatorname{sgn} x$
Since $x\operatorname{sgn} x = \left|x\right|$, this is $2\left|x\right|$.
$= 2\left|x\right|$

Answer

\left|x\right| + x \operatorname{sgn}\left(x\right)

No corner here — multiplying by x has ironed it out, and f′ is the V $2\left|x\right|$ touching zero smoothly at the origin.

Unusually, this one is differentiable at $0$: the product smooths the corner out, and the derivative there is $0$.

Open x abs(x) in the calculator

Example 4

\frac{d}{dx}\left[\left|x^{2} - 1\right|\right]

Chain rule with $u = x^{2} - 1$.
$= \operatorname{sgn}\left(x^{2} - 1\right)\cdot 2x$

Answer

2 x \operatorname{sgn}\left(x^{2} - 1\right)

A parabola whose middle has been reflected upwards into an arch, with f′ breaking at both folds.

Two corners, at $x = \pm 1$, where the parabola crosses the axis.

Open abs(x^2-1) in the calculator

Example 5

\frac{d}{dx}\left[\left|\sin\left(x\right)\right|\right]

Chain rule with $u = \sin x$.
$= \operatorname{sgn}\left(\sin x\right)\cos x$

Answer

\cos\left(x\right) \operatorname{sgn}\left(\sin\left(x\right)\right)

A row of identical arches sitting on the axis, with f′ jumping sign at the foot of every one.

A corner at every multiple of $\pi$ — the rectified sine wave.

Open abs(sin(x)) in the calculator

Watch out for

There is no derivative at the zeros of the argument, only on either side of them.
$\sqrt{x^{2}} = \left|x\right|$, not $x$, so its derivative is $\operatorname{sgn} x$.
$\operatorname{sgn}$ itself has derivative $0$ away from the origin — it is locally constant, not linear.

Chain rule Product rule Constant rule

What it says

When it applies

Five worked examples

Watch out for

Related rules