# Derivating absolutes, etc

1. Feb 11, 2005

### Alkatran

Is there some general formula for deriving an absolut-ed function? Is what I;m doing wrong (a lot of derivation relies on continuous functions, doesn't it?)

IE:
d/dx(abs(sin(x)))

Here's what I got:
abs(x) = x*sign(x)
d/dx(sign(x)) = 0 (x != 0)
therefore
$$\frac{d}{dx}abs(f(x)) = \frac{d}{dx}f(x)*sign(f(x)) = f '(x) * sign(f(x)) + 0$$

Which, by FTC would mean that:
$$\int cos(x)*sign(sin(x)) \dx = abs(sin(x))$$
'I checked this by drawing the graphs and it appears right...

Also... I saw that:
$$abs(sin(x)) = sin(x \mod \pi)$$
$$\int x \mod c \dx = (\int_{0}^{c} x \dx)*INT(\frac{x}{c}) + \int_{0}^{x \mod c} x \dx$$
example:
$$\int x \mod 1 \dx = INT(\frac{x}{c}) * .5 + x \mod 1$$
continuing...
$$\int abs(sin(x)) dx = \int sin(x \mod \pi) \dx = (\int_{0}^{pi} sin(x) dx)*INT(x / \pi) - cos(x \mod \pi) = 2*INT(\frac{x}{\pi}) - cos(x \mod \pi)$$
Far as I can tell it works...

Last edited: Feb 11, 2005
2. Feb 11, 2005

### AKG

Consider f(x) = x. To the right of zero, it's derivative is 1, and to the left, it is -1. The derivative, you can easily show, does not exist at 0. Do this from first principles, where you know the derivative is a limit. To find this limit, calculate the right limit (as your variable, normally h, approaches zero from the right) and notice that the limit evaluates to 1. Notice that it is -1 when h approaches zero from the left. Therefore, since left limit is not equal to right limit, the limit doesn't exist, and so, by definition, the derivative doesn't exist (since the derivative is this very limit).

3. Feb 11, 2005

### Alkatran

Sorry, I completely forgot to add x != 0

d/dx(abs(x)) = sign(x), x != 0