Question

Is it possible to give the derivative of |x| at x=0 a value?

Answer 1

no, the derivative is undefined at x = 0

Answer 2

the derivative as x approaches zero from the right is 1 , the left hand slope is -1

Answer 3

Yes the algorithm doesn't define a derivative there, but look at this graph:

https://www.desmos.com/calculator/cwbnewkrud

We can see that in order to turn through that single point the derivative there looks like it should, if it did have a value from the algorithm, take on a value between -1 and 1.

Not only that, it's at a local min, which normally means that the derivative should be zero there, which also lies between -1 and 1.

Just because it doesn't approach it doesn't mean we can't define the derivative to be 0 there since it seems to fit. It seems more useful than leaving it undefined and it makes sense to do so I think.

Answer 4

What algorithm?

Answer 5

This one: \[f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\]

Answer 6

for h>0 :
(f(0 + h ) - f(0)) / h = f(h)/h = h / h = 1 , and lim 1 = 1

for h <0
(f(0 + h ) - f(0)) / h = f(h)/h = -h / h = -1, and lim -1 = -1

Answer 7

Right, I'm not really debating that the algorithm doesn't define anything there, but I did give two good reasons why we should expand the concept to say the derivative should be 0 at x=0.

Answer 8

that limit does not exist at x = 0 , so the derivative is undefined.
$$ \large f'(x) = \lim_{h \to 0^{+}} \frac{f(x+h)-f(x)}{h}= \lim_{h \to 0^{-}} \frac{f(x+h)-f(x)}{h}


$$

Answer 9

the derivative must equal the right hand limit and the left hand limit if it is to exist

Answer 10

I don't think you're really paying attention to what I'm saying.

Answer 11

We have an accepted definition of derivative. I don't know why you want to violate it.

Answer 12

Your picture shows there isn't a unique tangent line at x=0.

Answer 13

^

Answer 14

Yeah it shows that at that point the slope could possibly be any value in a range from -1 to 1. But the slope is definitely not -3 or 7 there.

Imagine a particle flying along that path, it will be the tangent vector. Right when it gets to x=0 it will have to turn through an angle before it can continue flying straight. I'm saying that the regular derivative definition seems like it might be too simple or naive. What if we need something better? It seems to have a range there.

My other argument is that this is a local min, and so commonly we find the derivative is 0 at a local min and max, so why not take this geometric stance in addition to or separately to expand it?

At the very least, what makes everyone so sure that limit definition is the one and only true and final derivative definition that can never be improved upon? What's stopping us from potentially revising and generalizing it perhaps? I don't have the answer to this question but it feels to me like you all have ruled this possibility out -- why?

Answer 15

local max/min can occur when f'=0 or when f' dne

Answer 16

because of the definitions already here

we can make a new definition maybe

Answer 17

You can show that f'(0) cannot be defined.
Assume f'(0) exists
\[ \Large \text{Definition}
\\
\large{ f'(x) = \lim_{h \to 0^{+}} \frac{f(x+h)-f(x)}{h}= \lim_{h \to 0^{-}} \frac{f(x+h)-f(x)}{h}
\\~\\ f' (0) = \lim_{h \to 0^{+}} \frac{f(0+h)-f(0)}{h} = 1
\\ f' (0) = \lim_{h \to 0^{+}} \frac{f(0+h)-f(0)}{h} = -1
\\ \therefore
\\ 1 = -1
}
\]

Answer 18

Yeah, please don't argue this same thing again @jayzdd I understand completely what you're saying, it's just not addressing my issue.

Answer 19

You could make an alternate definition of derivative, but we have a lot of theorems that depend on the limit definition of derivative. So you would have to start from scratch in light of this new derivative.
In topology there is an other way to define derivative, so this is not the last word on derivative.

Answer 20

g star of a function g:

\[g^{*}(x)=g'(x) \text{ if } g'(x) \text{ exist } \\ g^{*}(x)=0 \text{ if } g'(x) \text{ doesn't exist } \]
so
g(x)=|x|
g star=x if x>0 and -x if x<0 and 0 if x=0


now why did we need this?