Derivative of Absolute Value Function
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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.
$\blacksquare$