1 Answer

Canucklesandwich · Answer 1 · 2023-03-05T03:13:29+0000

Recall the definition of absolute value.

$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$

When
x<0 ,

$|x| = x \implies (d|x|)/(dx) = 1$

When
x<0 ,

$|x| = -x \implies (d|x|)/(dx) = -1$

The derivative does not exist at
x=0 , since the one-sided limits

$\displaystyle \lim_(x\to0^-) f'(x) = -1$

and

$\displaystyle \lim_(x\to0^+) f'(x) = +1$

do not match.

So the derivative of
|x| is

$(d|x|)/(dx) = \begin{cases} 1 & \text{if } x > 0 \\ -1 & \text{if } x < 0 \\ \text{undefined} & \text{if } x = 0\end{cases}$

Now we can write this as

$(d|x|)/(dx) = \frac x = \fracxx$

since

$x > 0 \implies |x| = x \implies \fracxx = \frac xx = 1$

and

$x < 0 \implies |x| = -x \implies \fracxx = -\frac xx = -1$

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