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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
[tex]
\frac{d}{dx}abs(f(x)) = \frac{d}{dx}f(x)*sign(f(x)) = f '(x) * sign(f(x)) + 0
[/tex]
Which, by FTC would mean that:
[tex]
\int cos(x)*sign(sin(x)) \dx = abs(sin(x))
[/tex]
'I checked this by drawing the graphs and it appears right...
Also... I saw that:
[tex]
abs(sin(x)) = sin(x \mod \pi)
[/tex]
[tex]
\int x \mod c \dx = (\int_{0}^{c} x \dx)*INT(\frac{x}{c}) + \int_{0}^{x \mod c} x \dx
[/tex]
example:
[tex]
\int x \mod 1 \dx = INT(\frac{x}{c}) * .5 + x \mod 1
[/tex]
continuing...
[tex]
\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)
[/tex]
Far as I can tell it works...
IE:
d/dx(abs(sin(x)))
Here's what I got:
abs(x) = x*sign(x)
d/dx(sign(x)) = 0 (x != 0)
therefore
[tex]
\frac{d}{dx}abs(f(x)) = \frac{d}{dx}f(x)*sign(f(x)) = f '(x) * sign(f(x)) + 0
[/tex]
Which, by FTC would mean that:
[tex]
\int cos(x)*sign(sin(x)) \dx = abs(sin(x))
[/tex]
'I checked this by drawing the graphs and it appears right...
Also... I saw that:
[tex]
abs(sin(x)) = sin(x \mod \pi)
[/tex]
[tex]
\int x \mod c \dx = (\int_{0}^{c} x \dx)*INT(\frac{x}{c}) + \int_{0}^{x \mod c} x \dx
[/tex]
example:
[tex]
\int x \mod 1 \dx = INT(\frac{x}{c}) * .5 + x \mod 1
[/tex]
continuing...
[tex]
\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)
[/tex]
Far as I can tell it works...
Last edited: