# Derivative of Absolute Value Function

## Theorem

Let $\size x$ be the absolute value of $x$ for real $x$.

Then:

$\dfrac \d {\d x} \size x = \dfrac x {\size x}$

for $x \ne 0$.

At $x = 0$, $\size x$ is not differentiable.

### Corollary

Let $u$ be a differentiable real function of $x$.

Then:

$\dfrac \d {\d x} \size u = \dfrac u {\size u} \dfrac {\d u} {\d x}$

for $u \ne 0$.

At $u = 0$, $\size u$ is not differentiable.

## Proof

 $\ds \frac \d {\d x} \size x$ $=$ $\ds \frac \d {\d x} \sqrt{x^2}$ Square of Real Number is Non-Negative $\ds$ $=$ $\ds \frac \d {\d x} \paren {x^2}^{\frac 1 2}$ $\ds$ $=$ $\ds \frac 1 2 \paren {x^2}^{-\frac 1 2} \cdot 2 x$ Chain Rule for Derivatives $\ds$ $=$ $\ds \frac x {\sqrt{x^2} }$ $\ds$ $=$ $\ds \frac x {\size x}$

$\Box$

Now consider $x = 0$.

From the definition of derivative:

 $\ds \valueat {\dfrac {\d \size x} {\d x} } {x \mathop = 0}$ $=$ $\ds \lim_{x \mathop \to 0}\frac {\size x - 0} {x - 0}$ $\ds$ $=$ $\ds \begin {cases} \lim_{x \mathop \to 0^+} \dfrac x x & : x > 0 \\ \lim_{x \mathop \to 0^-} \dfrac {-x} x & : x < 0 \end {cases}$ Definition of Absolute Value $\ds$ $=$ $\ds \begin {cases} 1 & : x > 0 \\ -1 & : x < 0 \end{cases}$

From Limit iff Limits from Left and Right, the limit does not exist.

$\blacksquare$