Symmetric derivative of abs(x) at 0

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Have you ever felt sad that we do not have a derivative of abs(x) at 0?
If so, then try symmetric derivative!

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Symmetric derivative is linear. The symmetric derivative of a differentiable function is just the derivative. Hence the symmetric derivative of x is 1 and so the symmetric derivative of x + |x| will be 1 + 0 = 1.

jamiewalker

The symmetric derivative of an even function (defined on the reals) is always 0 at x=0.

jamiewalker

Fs(0) if f=x+abs(x)=1, just by applying the definition. You get the limit of 2h/2h which is 1

Catilu

I had seen sometimes that if f(x)=|x| then f'(x)=|x|/x. And yes, in this case f'(0) D.N.E.
But it isn't correct, in my opinion.
Actually, f'(0) exists because (|x|)'=sgn(x) and sgn(0)=0. Therefore, f'(0)=0.
Derivative of sum is the sum of derivatives.
f(x):
|x|=x if x>0,
|x|=0 if x=0,
|x|=-x if x<0;

f'(x):
x>0: x'=1;
x=0: 0'=0;
x<0: (-x)'=-1.

It's also sgn(x).

No limits were needed here at all.

renpo

The symmetric derivative is simply the arithmetic mean of left and right derivatives (that's why you get 0 for |x| because the left derivative is -1 for x<0 and the right derivative is 1 for x>0). Hence, for x+|x| you get (0+2)/2 = 1.

Kapomafioso

Coool!!!! never learned this in cal 1...clearly i have a lot to learn in calculus

buhle

This man loves math and math too loves him...for sure :) 😀

e-learningtutor

Is Fs(a) the mean of the right and left limits of f'(x) as x goes to a ? That was my intuition before watching the video 😉

beaming_sparkling_trash

if x<0 then f(x) = 0 and if x>0 then f(x) is 2x
f(0-h) = 0
f(0+h) = 2*0+2h
by definition of symmetrical derivative you get lim h->0 (2h - 0) / 2h which is 1

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