Question

How derivative of this absolute value function is like this!
Can someone give me step by step solution please ? ​

Answer 1

Recall the definition of absolute value.

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

When $x<0$,

$|x| = x \implies \dfrac{d|x|}{dx} = 1$

When $x<0$,

$|x| = -x \implies \dfrac{d|x|}{dx} = -1$

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

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

Now we can write this as

$\dfrac{d|x|}{dx} = \dfrac x{|x|} = \dfrac{|x|}x$

since

$x > 0 \implies |x| = x \implies \dfrac{|x|}x = \dfrac xx = 1$

and

$x < 0 \implies |x| = -x \implies \dfrac{|x|}x = -\dfrac xx = -1$