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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,657

hi guys

i was wondering what derivatives of |x| and sgn(x) are.i thought of something but i got that they were sgn(x) and 0 respectfully.i think something is wrong here.What?

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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,528

hi Stefy,

Haven't 'spoken' to you in a while. Hope you are well.

Derivatives have to be 'well defined'. Specifically, this breaks down where the left limit isn't the same as the right limit.

So neither have a defined value at x = 0.

For |x| the deriv = 1 for x > 0 and -1 for x < 0.

How are you defining sgn(x) ?

If you mean 1 when x > 0 and -1 when X < 0, then zero is correct.

As this function is discontinuous at x = 0, I think you must exclude x = 0 from this derivative even though the value is the same either side of zero.

Bob

*Last edited by bob bundy (2011-11-11 20:45:55)*

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,657

hi bob

yes,i have been occupied by school and my subjects.this was a crazy week.i'm glad it's over,but there is next week exams and i have physics, analysis and in the one after i have geometry and programming. very busy weeks!

anyway,if i gave the restriction x<>0 then would my answers be correct?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,528

hi Stefy

Yes, that's what I think.

Good luck in the exams (not that you need it .. I'm sure you'll do brilliantly)

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,657

ok thanks for the help,and could sure use luck!I'm not sure about that brilliantly!

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**reconsideryouranswer****Member**- Registered: 2011-05-11
- Posts: 172

anonimnystefy wrote:

i was wondering what derivatives of |x| and sgn(x) are.

i thought of something but i got that they were sgn(x) and 0

respectfully.i think something is wrong here.What?

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,657

nice explanation! thanks rya!

i guess sgn(x) can be done similairly.

Here lies the reader who will never open this book. He is forever dead.

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**zetafunc.****Guest**

reconsideryouranswer wrote:

anonimnystefy wrote:i was wondering what derivatives of |x| and sgn(x) are.

i thought of something but i got that they were sgn(x) and 0

respectfully.i think something is wrong here.What?

Sorry for the bump, but could you not just say

?

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,657

That looks ok too.

Here lies the reader who will never open this book. He is forever dead.

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