Derivative of absolute value function

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I love his energy as he teaches, it makes easier to focus on what he says

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

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Visited this channel 1st time in my life and now I'm in love with the way he teaches so calmly

[@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 :)

[@abdulmoeed581]

To the point, with perfect explanation, telling exactly what we need to know without stretching it way too long nor just doing the solutions! Subscribed

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Man, you are my type of teacher. You're so calm and easy catch. Keep doing what you do.

[@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.

[@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

[@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.

[@edgardojaviercanu4740]

The key is the deffinition of the absolute value. A fine work, indeed!

[@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

[@AaaAaa-yb2nb]

Omg 😭 i had no idea you could write absolute value like that, thanks so much, now I can really handle these problems

[@xxasian00bxx44]

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

[@HaleyNgo-hi1pv]

this video is amazing! i learned so much in such a short amount of time! thank you sir

[@FredDeliege]

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

[@LinasDanusevičius]

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

[@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.

[@arefhamaoui9962]

You explained the concept perfectly, thank you sir.

[@bevansomondi6399]

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

[@AndiswaManqele]

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

[@neelkhatri4793]

You're an absolute mathematical dude man!

[@PrimeNewtons]

Lol. I see the pun.

[@solipse.]

Interestingly, you can flip the absolute of the function and the function (numerator and denominator) and it still works. we want the derivative to divide by zero when the function is 0. and the function divide itself with the modulus one producing “1” * the derivative of function. then because one of them is modulus and function is flip, so the derivative is also flip on the x axis, when the function was increasing will be the negative of derivative of those interval, we will retain the negative sign of function but not from the modulus producing the flip we want in those interval which makes it correct. pretty neat.

[@felipesants8936]

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

[@dontforgetlungi]

Great explanation!!

[@anshumanrawal2115]

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

[@corazonkkk]

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

[@Umairhassan279]

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

[@PrimeNewtons]

Haha. Thank you.

[@punditgi]

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

[@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.

[@ERA.P01]

Best one, great mr, 💯💯🎉🥳

[@mihnwq9211]

Really good if you don't want to split the function into two parts as in your example , based on the fact if x >= 3/2

[@RT1649]

You earned a Subscriber ❤ Great Teaching Methodology, Keep Going

[@advaithrvasistha44]

earned a subscriber ,amazing teaching

[@anshumanrawal2115]

Bro u are so good in maths ❤❤❤❤❤

[@OS-ph6sk]

Awesome video and explanation, genuinely love your content!

[@elliotappiah7434]

Thanks a lot 🤲😊

[@dontstudymath]

awesome class man! For a moment I felt like Kamaru Usman from UFC was teaching me a lesson xD. Joke aside, neat explanation.

[@bloodghoul1129]

than you so much sir i just couldent get any idea of chapter i just thought maths just isent my cup of tea but i got a hang of it thanks to you

[@TorgawonBenjamin]

Sir u are too much I learned a lot from you

[@Christian_Martel]

The advantage of writing y = |x|, and y’ = x/|x| is that clearly shows that y’ is undefined at critical point x=0.

[@slavinojunepri7648]

This is very cool. For completeness, the derivative of |x| does not exists at x=0. By extention, the derivative of |2x-3| doesn't exist at x=3/2.

[@rasharign.]

Here from 🇮🇳 india 🇮🇳 ❤

[@davyfoad7264]

thanks man, you helped me on my test

[@sandrunner9013]

Man this was incredibly helpful! Thank you so much!

[@PrimeNewtons]

Glad it helped!

[@ChalaFokora]

Wow! I have no word to express you.

[@Hiraeth256]

Beautifully done. Thank you mate.

[@kevinkasp]

I love your channel so much.

[@PrimeNewtons]

Thank you so much!

[@soun-jawalters1773]

Excellent explanation. Subscribed.

[@Matiullah-mm8lv]

🎉 yes oky please explain common shapes of absolut function

[@mabenj6954]

This guy is so good

[@mabenj6954]

(no homo tho)

[@jumpman8282]

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

[@AbdullahMuzaffar-z2t]

very good explanation

[@Jason-ot6jv]

Great video! keep up the great work!

[@RandaAltajee]

saved my life thank you so much

[@robert19]

legend 💪 keep it up man!!

[@richardrobertson1886]

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

[@Sooha20]

Amazing video ❤

[@aashsyed1277]

5:48 the chain rule can also be used

[@arunprasathgm]

Excellent

[@AliHaider-j1d3f]

Really helpful

[@patrick-matematicalda4209]

From Angola 🇦🇴

[@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.

[@cowboyclay8783]

thank you for the help sir

[@adinathpattathil7592]

Wonderful video. Thanks!

[@antling_]

great video

[@WakeBikila]

Tnx professor

[@surendrakverma555]

Very good. Thanks 👍

[@onlqcngel]

that was really helpful, thank you!

[@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.

[@bartomiejosiak3287]

Great video!

[@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.

[@christiesfeir7215]

Super helpful!

[@joelgerard7869]

Good stuff!

[@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.

[@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

[@Jfsc]

love, thanks brother

[@wsellias7360]

Thankyou! this was so helpfull! :D

[@capybara-k6g]

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

[@ayanahmed5114]

can you do another epsilon delta proof

[@begamsahanajsultana1676]

very good vi deo

[@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

[@davidhanmin]

Thank you!!

[@rob876]

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

[@slavinojunepri7648]

Well, sign(0) does not exist.

[@cameron_graham95]

Brilliant😁

[@AmandaGumede-qv2cl]

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

[@ylitz]

Thank you thank you thank you

[@albajasadur2694]

Why don't we split the absolute value x into three ranges, x < 0, x=0, and x> 0 and do the differentiation in each range? The derivative expression x/|x| is not defined at x=0, implying abs(x) is not differentiable at x=0.

[@ramziyasebargaeva]

👍👍👍

[@meow11119]

AMAZING

[@commonsensedude6211]

“The absolute value is always a V”

[@Maraquitica]

Te amo.

[@costelnica3988]

Se putea rezolva si prin explicitarea modulului?

[@TylerRich-bw9zh]

helpful ty bro

[@SachinSharp]

lifesaver

[@i_understand_NOTHING_v]

Nice vid

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Legend

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❤❤

[@ayo509]

thanks

[@PrimeNewtons]

You're welcome!