Question

How do you calculate the derivative of an absolute value, e.g. g' if g(x) = |x - 2|

Answer 1

its similar to calculating: sqrt(x^2)

Answer 2

It all boils down to the fact that absolute value is basically just a piecewise function in disguise.
In particular
\[\Large g(x) = |x-2|= \left\{\begin{matrix}x-2 &&& x\ge2\\\\2-x&&&x<2\end{matrix}\right.\]

Answer 3

I saw that (square root of a square) in a Google search but my study guide shows something else. I think I'll ask this again once I get home to check my study guide.

Answer 4

\[\sqrt{x^2}~:~let~u=x^2\]
\[\sqrt{u}\to\frac{u'}{2\sqrt{u}}\to\frac{2x}{2\sqrt{x^2}}\to\frac{x}{|x|}\]

Answer 5

oh....^
Das geht... lol

Answer 6

:)

Answer 7

@terenzreignz Love the way you explained that. It actually makes learning absolute value more fun. "...in disguise..". lol.

Answer 8

Yeah... most unfortunately, it'd probably be a headache if it were more than just the absolute value of some linear function... imagine polynomials... we'd have to find its roots D:

Answer 9

hehehehehe @terenzreignz But if you know the basic definition, no matter how difficult the problem, you'd be able to figure it out =]

Answer 10

Any time now someone can show me how to do a derivate of |x-2| lol. Bear in mind that I only started calculus a few weeks ago and we didn't have calculus at high school.

Answer 11

@Mandre Now according to the piecewise function @terenzreignz gave you, differentiate each and see what you get. Each derivative will correspond to the respective interval.

Answer 12

And you'll see that it jives (fortunately) with amistre's more elegant derivative :)

Answer 13

\[\Large \frac{d}{dx}g(x) = \frac{d}{dx}|x-2|= \left\{\begin{matrix}\frac{d}{dx}[x-2] &&& x\color{red}{>}2\\\\\\\\\frac{d}{dx}[2-x]&&&x<2\end{matrix}\right.\]

Answer 14

Then it's 1 for (x-2) or -1 for (2-x)?

Answer 15

why not? Something to do with continuity?

Answer 16

@Mandre
That is correct :)
Notice I replaced the \(\ge\) sign in the top-function of the piecewise with a simple > sign...
That is because the derivative simply doesn't exist at the point x = 2.

Answer 17

I need to revise my continuity. It seemed so simple at the time * sigh *

Answer 18

my derivative is undefined for x=2 ...

Answer 19

@amistre64 how is it undefined at '2'?

Answer 20

@amistre64
\[\Large \frac{x-2}{|x-2|}\]?

Answer 21

Div by 0

Answer 22

\[\frac d{dx}|x-2|\to\frac{x-2}{|x-2|}\]

Answer 23

since |x-2| has no derivative at x=2 to begin with ....

Answer 24

Oh it's (x-2). Sorry I forgot that lol =P

Answer 25

my initial post was really about a linear absolute value in general

Answer 26

@amistre64 oh ok lol. I thought that it was the question rofl...

Answer 27

i had made an explicit rule for some sequence in class .... using the derivative of a combination of linear absolute values such that at any given "n", it matched the stated sequence .... over kill, but it helped to stem the tide of boredom

Answer 28

Here is my actual problem. Find the derivative of g
\[g(z) = \frac{ z ^{3}-z+1 }{|1-2z| }\]

Answer 29

I know how to use the quotient rule but get unstuck at the abs value.

Answer 30

you could either divvy it up into parts like terens did, or use my demonstration

Answer 31

Oh, lol... we can't have piecewise functions, now can we? :P
You better use amistre's derivative...

Answer 32

\[g(z) = \frac{ z ^{3}-z+1 }{(1-2z) };~\frac{ z ^{3}-z+1 }{(2z-1) }\]

Answer 33

Thank you amistre. You're a star.

Answer 34

Thanks everyone

Answer 35

yep, that me ... the lorenzo lamas of openstudy lol

Answer 36

lol.