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Airida [17]
1 year ago
8

How derivative of this absolute value function is like this!

Mathematics
1 answer:
Daniel [21]1 year ago
6 0

Recall the definition of absolute value.

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

When x,

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

When x,

|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

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