Derivative of Absolute Value Confusion

[russ_watters]

Quick question (hopefully): I'm helping my girlfriend with some calculus homework over the phone and it looks like the book has a different answer for the derivative of an absolute value than I'm seeing in other places. I'm seeing it as (and I derived it myself to be) x/abs(x) whereas the book (apparently) has abs(x)/x. But it seems to me that these should be equal. Are they?

Why this matters is if you try to apply this to problems, you get some very ugly things.

What am I missing?

 

[Office_Shredder]

x²=|x|²
x*x = |x|*|x|
x/|x| = |x|/x

So yes these things are equal. The ugly things might be coming up because the function is not differentiable at 0 and of course neither of these is defined at zero. If your problem involves crossing x=0 weird things can happen. Can you give an example?

 

[russ_watters]

Given: y=|x|+1/x
Solution: y=(x²)^1/2+x^-1
y'=1/2(x²)^-1/22x-x^-2
y'=x/|x|-1/x²
Can also be written: y'= (x|x|-1)/x²

But unless I'm mishearing her, the solutions manual says:
y'=|x|/(x-1/x²)
...and they don't look equal to me.

 

[Gokul43201]

Maybe it's just a typo (or a mishear), intended to be y'=|x|/x - 1/x².

 

[russ_watters]

Yeah, I asked several times, but it is possible that it is either.

 

[russ_watters]

Yeah, I got her on a webcam and it was an algebra comprehension issue. That's what I thought but I wanted to make sure I wasn't crazy.

 

[Drinknderive]

y = |x|
implies y = ax and y = -ax
therefore y' = a and y' = -a