Derivative of |x| (absolute value x)

In this short video I show how to find the derivative of |x|, the absolute value of x. In one minute, I cover the square root definition of absolute value x, the Chain Rule and the Power Rule. This formula is hard to remember in calculus, so I am teaching how to derive the formula so that you will never forget the derivative of absolute value x.

[dmr]

I have to say, as a mathematician I think if one of my students saw d/dx of a piecewise-linear function and did this derivation, I’d give them the points but instruct them in the future to solve this one geometrically. I think most of the time, reducing a problem to known results is a more valuable and transferable skill than algebraic manipulation, so I’d want them to first notice the non-differentiable point at the origin, then see that on the rest of the domain the function is either x or -x

[channalbert]

This is a very good derivation, but you should've mentioned that the domain is reduced to R-{0}.

For anybody wanting more, there is a branch of mathematics called distribution theory in which this derivative would be equal to the same as in here plus a "new" generalized function, or distribution, called Dirac delta, centered at 0 with a factor of two.

[YRO.]

I've never heard of absx being the same as sqrtx^2, that's crazy

[oroboros]

Excellent derivation sir, the derivative also shows non-differentiable at x=0 which makes sense since abs(x) has a cusp at x=0

[jase]

this random video just opened up sooo much equation on my head

[julianbruns]

|x| isn't differentiable at x=0 but it has the weak derivative sgn(x) (or any function that equals sgn(x) almost everywhere sincs those get identified in this context) on all of R.

[randvar]

x/|x| * 1_{x<>0} (The calculation starts with x over all reals, but it ends up with x excluding 0. The drop appears in the derivative of the power composition. It’s all good except the third equality sign which needs a stackrel{x<>0}.)

[Urshrek]

My question is can we write the final answer to be 1 or -1? If not then why?

[nomzz]

Another way you can do this derivative is with implicit differentiation. Define y = |x| and therefore y^2 = x^2

Taking derivatives of both sides gives 2y dy/dx = 2x

Dividing by y on both sides gives dy/dx = x/y

And then from original definition, y = |x|, so dy/dx = x/|x|

On a side note, this is also known as the sign function, sgn(x), although at sgn(0) = 0 whereas in this case it’s undefined.

[JJ_TheGreat]

Question:
Who says that the absolute value of x is equal to the square root of x²?

[zkprintf]

It's much simpler to apply |x|=signum(x)×x. Signum has an isolated point at 0 and is piecewise constant everywhere else.

[GrZeNiOFantasy]

So basically step function with undefined 0

[jamiewalker]

|x| = sgn(x) x which differentiates to give sgn(x) when x isn't 0. If you really want, you can write sign(x) as x/|x|

[perseusgeorgiadis]

You should mention the function is non-differentiable at 0 (which makes sense since you have |x| in the denominator too)

[fleurpayne]

thats so sick!!! i do need to fall asleep now and this was very relaxing ❤

[murcury]

It’s not a derivable function as the is a shaft point where the limits of the derivative on rather side won’t match up

[JustFiscus]

So it’s just equal to 1 for all values of x not equal to zero right

[amichayr]

I think it's pretty weird to not go one step further and say 1 if positive, and negative one if negative

[talenmud]

Me seeing this straight after my teacher taught us the bracket derivative rule

[dineshmanaria]

But |x| is not differentiable function how you can commentbon Diferentiability