Derivative of x+abs(x)

We will find the derivative of f(x)=x+abs(x). We will use the piece-wise definition of abs(x). This is another continuous function but not differentiable at x=0.

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#blackpenredpen #math #calculus #apcalculus

[Apollorion]

Mmm... considering this function, would you rather derive a: it's _antiderivative, _ b: it's _primitive function, _ c: it's _primitive integral_ or d: it's _indefinite integral_ and *why* ?

[jackiekwan]

#yay
Finally there's one video that I can find out the answer on my own haha
I DO know this is an easy one though...

[angelmendez-rivera]

Actually, if we use the Dirac delta distribution δ(x), then we can define the derivative of the function at x = 0. Notice that |x| = x*sign(x), where sign(x) is the function which is -1 when x < 0, 0 when x = 0, and 1 when x > 0. Then f(x) = x + |x| = (1 + sign(x))*x. Therefore, f’(x) = x*d sign x/dx + 1 + sign (x). Given the definition of δ(x), d sign x/dx = 2 δ(x), hence f’(x) = 2x*δ(x) + 1 + sign(x). x*δ(x) = 0 is a derived property of δ(x), thus f’(x) = 1 + sign(x), so f’(0) = 1 + sign(0) = 1.

[matthewstevens]

It isnt piecewise if you just define abs(x) as sqrt(x^2). you can differentiate that to get 2x(1/2sqrt(x^2)) = 2x/2abs(x) = x/abs(x). Adding the 1 from derivative of X you have the expression 1+ x/abs(x) for the derivative - gives 2 for positive values, 0 for negative and undefined for 0. This is due to the X/abs(x) giving out the sign (+/-) of the input multiplied by 1, hence this result

[medpop]

I really love the way you use to explain

[MrRyanroberson]

Take sqrt((sqrt(x)+sqrt(y))^2) to find the identity for adding the radicals = sqrt(x+2sqrt(xy)+y), the same applies for x+y=sqrt(xx+2xy+yy)
So I then tried adding sine and cosine this way to find that...
Sin+cos=sqrt(1+2sin(x)cos(x))
And as it turns out, this function behaves a bit funny when you subtract the sqrt term from both sides and set the whole thing equal to y and plot it as a graph. The result resembles either a positive or inverted version of what the solar power function looks like at the equator.

[AngelDemon]

Should the derivative not be x/|x|+1, as per wolfram alpha? I realize that the solution is the same when you plug in the numbers at x>0, and x<0, but isn’t this answer more general since it uses the derivative of |x|=x\|x|?

[dutchkaluuk]

cant you turn the picewise function back to the normal function f(x)=(x + abs(x))/x

[michalbotor]

why stop here? the real fun was just about to begin! ;)

[miguelyanac]

Excellent video. I see it from Córdoba (Argentina). A greeting. (y)

[michel_dutch]

I would love a vid about the Dirac Delta function.

[wolswinkel]

would have been fun to go back to a non-piece wise definition for the derivative, like x/|x|+1 #YAY

[benjaminbrady]

It is possible for a discontinuous function to have a continuous derivative 🤔

[ralfbodemann]

Next video is about the second derivative of f(x)=x+abs(x), esp. at x=0.
BTW: I'm a physicist.... ;)

[pablojulianjimenezcano]

All your videos are great!! Nice "corner" function ^-^
#Yay #BulliedCalculusTshirt

[larsmarz]

#YAY Keep those vids coming, but without the blue pen next time ;)
Greeting from Germany :)

[romanhredil]

#yay did you know that d/dx(|x|)=|x|/x?
i'll show you the proof:

|x|={x if x>0, -x if x<0, 0 if x=0}
If you analyze the derivative of |x|:
d/dx(|x|)={1 if x>0, -1 if x<0, div. if x=0}
And if you analyze the function |x|/x:
|x|/x={1 if x>0, -1 if x<0, div. if x=0}

As you can see, these functions are the same.

[JMS]

f'(x) = 1 + x/abs(x) (for x=/= 0) :D

[jamieee]

the derivative is |x|/x + 1 #YAY

[deoxal]

Thought you were going to talk about symmetric derivative...disappointed.