Derivative of absolute value function

In this video, I showed how differentiate an absolute value function

[@leminator13]

Man, this guy is better than any lecturer I've ever had. Great youtube channel

[@user-zl3kl6vb1k]

Your videos definitely deserve much more attention, I feel lucky that youtube reccomened me your channel

[@josephparrish7625]

I always started my calc and precalc classes by teaching the definition of absolute value and how to use it. Absolute value is an easy concept but tricky if you don’t know how to apply it. Great lesson!

[@PrimeNewtons]

That's good practice for everyone

[@quest4knowledge-xh3eu]

This stuff is tricky :)

[@ernie345]

I'm stunned by how you explained the concept. You got a new subscriber

[@dottemar6597]

Good job. x/|x| also called Signum function (sgn(x)), so applying chain rule produces then dy/dx = 2sgn(2x-3), (x ≠ 3/2). I had the fortune of having a really good Calc book (Adams) when I studied back 20 years ago that hammered down certain basics in the very first chapter, like Heavyside and Signum.

[@jsproduction24]

thank you for your effort, it is so clear and helpful!

[@xxasian00bxx44]

came looking for an answer to my question and the explanation answered before the video started. thank you!

[@Hiraeth256]

Beautifully done. Thank you mate.

[@user-pe2yh6nq8s]

I like the energy of this video. Very good explanation. Thank you

[@OS-ph6sk]

Awesome video and explanation, genuinely love your content!

[@felipesants8936]

Thank you Sir for the great explanation. Never stop learning . All the best to you

[@oSilly]

Hello! Just wanted to say that your lessons are very well structured! I love how you take the time to explain each little step, making sure we understand the process to achieve the solution, instead of just giving the quick way to solve the derivative of an absolute value immediately. I'll definitely be watching your videos for any other topics I'm confused in. Thank you :)

[@PrimeNewtons]

Thank you for the feedback. Hope to keep improving every day

[@jacklion109]

This is a real g video. You explain everything so well! As an American student in Calculus BC who forgot this topic, this video was an amazing review. Thank you so much sir.

[@Jason-ot6jv]

Great video! keep up the great work!

[@user-su3ey7ih4x]

you and JG are the best i've ever seen on youtube

[@advaithrvasistha44]

earned a subscriber ,amazing teaching

[@user-wh2kn1bp9q]

Great explanation . With your help you help a lot of people struggling with math...no doubt. Cheers from Belgium

[@sandrunner9013]

Man this was incredibly helpful! Thank you so much!

[@PrimeNewtons]

Glad it helped!

[@anshumanrawal2115]

Never ever get demotivated brother remember a person from india will always be your no 1 supporter❤❤❤ brother injust subscribed❤

[@robert19]

legend 💪 keep it up man!!

[@anshumanrawal2115]

Bro u are so good in maths ❤❤❤❤❤

[@davyfoad7264]

thanks man, you helped me on my test

[@adinathpattathil7592]

Wonderful video. Thanks!

[@ERA.P01]

Best one, great mr, 💯💯🎉🥳

[@Sooha20]

Amazing video ❤

[@ZmQuad]

Thank you very much for this video. i have been struggling for two days, to differentiate the following: (x)ln|x|. i kept geting ln|x|+1, and ln|x|-1, but the calculator i use, was only evaluating the ln|x|+1. i could not figure out why, and this explains it. Thank you very much.

[@soun-jawalters1773]

Excellent explanation. Subscribed.

[@user-rm1xh7ob1j]

Wow! I have no word to express you.

[@bevansomondi6399]

He got it in him, a way of making maths looking too easy, fun and enjoyable.

[@neelkhatri4793]

You're an absolute mathematical dude man!

[@PrimeNewtons]

Lol. I see the pun.

[@kevinkasp]

I love your channel so much.

[@PrimeNewtons]

Thank you so much!

[@RandaAltajee]

saved my life thank you so much

[@christiesfeir7215]

Super helpful!

[@antling_]

great video

[@gganpatiprasad9112]

Thank you sir, you are a wonderful teacher. Love from India. Never stop learning.... Please complete the other sentences uttered by you at the end of your class.

[@punditgi]

The absolute truth is Prime Newtons is number one! 🎉😊

[@cowboyclay8783]

thank you for the help sir

[@davidhanmin]

Thank you!!

[@123qopsiznoq]

AMAZING

[@phiefer3]

Another approach, that gives you the same solution, and is arguably easier is to use a piecewise definition for the absolute value, then you can differentiate each piece individually. IE |x| = x when x>0, or -x when x0 and -1 when x

[@bartomiejosiak3287]

Great video!

[@TylerRich-bw9zh]

helpful ty bro

[@mudasirmalik8053]

Dear your personality and also your acting skills are amazing you should be in Hollywood movies.

[@PrimeNewtons]

Haha. Thank you.

[@surendrakverma555]

Very good. Thanks 👍

[@joelgerard7869]

Good stuff!

[@ayanahmed5114]

can you do another epsilon delta proof

[@Matiullah-mm8lv]

🎉 yes oky please explain common shapes of absolut function

[@mabenj6954]

This guy is so good

[@mabenj6954]

(no homo tho)

[@cherryang4644]

is it the same for || x ||, norms of a vector?

[@Emine-ri7ex]

it's very interesting video but how can we find the second derivative of absolute value function to determine the inflection point, concavity, and related character of the function?

[@PrimeNewtons]

I don't know

[@Emine-ri7ex]

@@PrimeNewtons what it mean

[@Noname61574]

i assume you can implement quotient rule and chain rule to get an answer

[@ant.pac7]

I would like to correct something if i may: mod(x) is actually equal to (√x)² and not √(x²). Square root rule states that the polynomial/variable inside any sqrt must be positive or zero. Since x² is always positive, you don't have to put mod around the x in √(x²) but x can be negative in (√x)² which would be undefined.

[@josephhassen]

I just took app. Maths exam 5hrs ago the last question was to differentiate this function and find the absolute extreme values. Unfortunately I missed it because I didn't know how it worked then. Now that I know it, I'm very disappointed for losing a 3pts. question

[@PrimeNewtons]

Aww. I'm sorry. You are better equipped for next time.

[@jumpman8282]

When you learn something you basically already knew, but never thought about. I feel so terribly _inside_ the box right now.

[@cameron_graham95]

Brilliant😁

[@patrick-matematicalda4209]

From Angola 🇦🇴

[@random_Person347]

I was surprised that you didn't go into what happens to the derivative when the expression goes from negative to positive. In the first example, differentiating y=|x|, the derivative is -1 when x0, but when x=0 it must be undefined, in other words y=|x| is not differentiable when x=0.

[@klara9314]

but what if we have a number multiplied by the absolute value of a function? can we find the derivate of the absolute value separately (like you did here) and just multiply it at the end with the number we have? please if u can answer me becaues too soon i have an exam 😫

[@Melmetal71]

Nice vid

[@atzuras]

'u' substitution seems a bit unnecessary since the chain rule works fine and it is often introduced earlier in the class. Fine proof anyway

[@Maraquitica]

Te amo.

[@ramziyasebargaeva]

👍👍👍

[@wasdc]

i never heard absolute value of x is = sqrt x^2. is this a definition that is true in all cases or did you just make an assumption here? thanks

[@PrimeNewtons]

For all real x, the square root of x² is +x or -x. Which is |x|

[@otuenufrancis1518]

Sir can you do a video on hw to evaluate square root limits eg limx~0(√x^2+9)-(2x^2+3/x^2))

[@PrimeNewtons]

Search in my channel. I already have many videos on that

[@williepham9562]

So can we generate a formular of derivative of an absolute function like this: f' (|x|) = |f'(x)| ?

[@costelnica3988]

Se putea rezolva si prin explicitarea modulului?

[@richardrobertson1886]

Wouldn’t it be easier to just view the absolute value function as a piecewise function?

[@MetalFaceDOOM3141]

I thought the dx of abs(x) is the sign function and therefore abs(x)/x?

[@MetalFaceDOOM3141]

I just realised it does not make a difference😂

[@AmandaGumede-qv2cl]

im sorry,but ive been avoiding this, yhooo you r so handsome

[@TheDuckTeamYT]

so what if you just multiply the exponents at (x^(2))^1/2 which is x^1 and now you have x = | x | which is not true.

[@googlegogel2673]

Simplifying a function changes it. f(x)=sqrt(x^2) and f(x)=x^1 is not the same function.

[@ayo509]

thanks

[@PrimeNewtons]

You're welcome!

[@evwerenisaacoghenenyerhovw232]

Sir please I need videos on partial differentiation please sir help me...... Multivariable calculus

[@isidorolorenzo802]

You should've taken into account the exceptions for each and every function, that is to say, X=0 & X=3/2, respectively, since the functions don't admit derivative for these values.

[@KipIngram]

Well, this is equal to 2x-3 for x > 1.5, and -(2x-3) for x < 1.5. So the derivative is 2 for x > 1.5, and -2 for x < 1.5. It's undefined at x = 1.5.

[@begamsahanajsultana1676]

very good vi deo

[@iikpp4]

خوش رجال

[@Nyambenyambe-xc9ib]

I don't follow what is happening here, I kept replaying the video but nothing to understand I just gave up.

[@Emine-ri7ex]

I'm waiting your response

[@rob876]

x/|x| = sign(x) = -1 when x < 0, 1 when x >= 0

[@sadeqirfan5582]

Couldn’t you just differentiate it piece-wise? Derivative will be piece-wise but won’t be defined at 0.

[@motivationmastery5442]

Any jee aspirant here ? 😅

[@niteshmayatwal]

Yes

[@user-ud1zv2yh3r]

I have never seen anyone make trivial math so complicated.