Derivative of an absolute value? (1 Viewer)

[kpq_sniper017]

What's d/dx of |x|?

The question:
f(x) = sin^-12|x|
Find f'(x).

???

 

[George W. Bush]

1 if x>0
-1 if x<0
not differentiable at x=0.

 

[jesshika]

umm from what i remember there will be 2 solutions
1 positive and 1 negative ..


that sounds like a question from the fitzpatrick book (26c or sumthing)

 

[eggnog]

d/dx {|f(x)|} = f'(x) * f(x) / abs(f(x))

the f(x) / abs(f(x)) takes into account the sign change

so d/dx {|x|} = x / |x|

 

[kpq_sniper017]

Originally posted by jesshika
umm from what i remember there will be 2 solutions
1 positive and 1 negative ..


that sounds like a question from the fitzpatrick book (26c or sumthing)

yep, exactly right. ex 26(c) - can't remember the question though.

anyone else also doing this exercise at the moment?

 

[grimreaper]

I believe that at x = 0 in your case, the gradient doesnt exist (something to remember)

 

[George W. Bush]

it's not like my first post didn't summarise this whole thread [except eggnog, who made a good contribution, and the whole fitzpatrick exercise thing]


oh btw pcx, the gradient is 1 for x>0. I'm not sure if I mentioned this before.

oh yeah, and the gradient is -1 for x<0.

 

[kpq_sniper017]

lol.
ok, i see now.

and if you had, for example d/dx |x² - 1|:
there'd be two solutions??
1. -2x
2. 2x

??

 

[George W. Bush]

Well, for x>1 or x<-1, f(x) = x^2 -1, so f'(x) = 2x
Now, for -1<x<1, f(x) = 1-x^2, so f'(x) = -2x

You also should state it's not differentiable at x=1, -1

 

[Affinity]

hmm...

as mentioned above the derivative of |x| is sgn(x) = x/|x|

so you just apply chain rule:

d/dx |x^2 -1| = sgn(x^2 - 1) * (2x) = (2x)*|x^2 - 1|/(x^2 -1)

 

[Xayma]

Affinity what does sgn mean? sign?

 

[Affinity]

yep