The antiderivative of the absolute value is apparently x|x| / 2, but when you differentiate that you don't get |x|?

Question

Unless I'm making a mistake (which I probably am), when you differentiate (x * |x|) / 2, you don't get |x|, which means (x * |x|) / 2 isn't an antiderivative of |x|?

Steps for differentiating (x |x|) / 2:

First, note that the derivative of (|x| / 2) is x / (2 |x|). I got this just by using the fact that |x| = (x^2)^1/2 and applying the chain rule.

Split it up: (x * |x|) / 2 = (x / 2) * (|x| / 2)

Apply the product rule: ( 1 * (|x| / 2) ) + ( (x / 2) * (x / 2 |x|) )

= (|x| * 2) + (x^2 / 4|x|)

So I'm getting the derivative as (|x| * 2) + (x^2 / 4|x|) ... where's my mistake? I think I differentiated everything correctly.

Any help is greatly appreciated.

Answer 1

It happens that$$\frac{x\lvert x\rvert}2=x\times\frac{\lvert x\rvert}2;$$you got the $\frac12$ part in both factors, which is wrong.

Answer 2

Consider what the product of $x$ and $|x|$ should look like. You'd get a piecewise function which is continuous. It is $-x^2$ for $x<0$ and $x^2$ for $x>0$.

Differentiate this along those intervals. You should get $2|x|$.

Answer 3

False statement in your argument: $$ \frac{x|x|}{2}=\frac{x}{2}\frac{|x|}{2} $$

Get rid of the factor $1/2$, but note that you cannot apply the product rule at $0$, because $|x|$ is not differentiable there.

With the product rule you have, for $x\ne0$, $$ F'(x)=\frac{1}{2}(x|x|)'=\frac{1}{2}\left(|x|+x\frac{|x|}{x}\right)=\frac{1}{2}2|x|=|x| $$

Now we also have to compute $F'(0)$: $$ F'(0)=\lim_{x\to0}\frac{F(x)-F(0)}{x}=\lim_{x\to0}\frac{x|x|/2}{x}= \lim_{x\to0}\frac{|x|}{2}=0 $$ Hence $F'(x)=|x|$ for every $x$.