Re: Absolute Value Differentiation

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Posted by Adam Getchell on October 04, 2002 at 00:51:53:

In Reply to: Re: Absolute Value Differentiation posted by Mary on October 03, 2002 at 20:41:59:

: : : Hi,

: : : How would I go about finding "for which values of X is the absolute value function y = abs x differentiate?"
: : :
: : : I am not sure either. But, if you plot y=abs(x) you get two lines, y=-x and y=+x. y=-x has a derivative over the range -infinity: : : I'm not sure where to begin. Perhaps I've forgotten too much about absolute values....

: : : Please help. Thanks!

:
: I found the formula for it, but I don't understand it.

: y(x) = |x|
: |x| = sqrt x^2
: x/dx |x| = d/dx *2sqrt(x^2)
: x/ |x|

: Therefore
: 1 if x >0
: -1 if x <0

: Does that make sense to you? I can't make sense of it.

Hmmm.

Whenever you solve an absolute value equation, you split it into the + and - parts. So |x| becomes
+x for x>0 and -x for x<0. Now take the derivative separately, and you get +1 for x>0 and -1 for x<0.

The way they're doing it above is expressing |x| as the square root of x squared. You set y = |x|, then y = sqrt(x^2), then square both sides to get y^2 = x^2. Now differentiate both sides to get 2ydy = 2xdx. You're really solving for dy/dx, so you now have dy/dx = x/y. But y = |x|, so you have dy/dx = x/|x|. Now let |x| = +x for x>0 and |x|=-x for x<0, and you get the same result.

Hope that helps!

Follow Ups:

you all are really making this too difficult... T.Gracken 07:50:18 10/04/02 (1)
- Re: you all are really making this too difficult... Adam Getchell 04:00:19 10/13/02 (0)

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: : : : Hi, : : : : How would I go about finding "for which values of X is the absolute value function y = abs x differentiate?" : : : : : : : : I am not sure either. But, if you plot y=abs(x) you get two lines, y=-x and y=+x. y=-x has a derivative over the range -infinity<x<0 and y=x has a derivative over the range 0<x<+infinity. : : : : I'm not sure where to begin. Perhaps I've forgotten too much about absolute values.... : : : : Please help. Thanks! : : : : I found the formula for it, but I don't understand it. : : y(x) = |x| : : |x| = sqrt x^2 : : x/dx |x| = d/dx *2sqrt(x^2) : : x/ |x| : : Therefore : : 1 if x >0 : : -1 if x <0 : : Does that make sense to you? I can't make sense of it. : Hmmm. : Whenever you solve an absolute value equation, you split it into the + and - parts. So |x| becomes : +x for x>0 and -x for x<0. Now take the derivative separately, and you get +1 for x>0 and -1 for x<0. : The way they're doing it above is expressing |x| as the square root of x squared. You set y = |x|, then y = sqrt(x^2), then square both sides to get y^2 = x^2. Now differentiate both sides to get 2ydy = 2xdx. You're really solving for dy/dx, so you now have dy/dx = x/y. But y = |x|, so you have dy/dx = x/|x|. Now let |x| = +x for x>0 and |x|=-x for x<0, and you get the same result. : Hope that helps!

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