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

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

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

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

10 ай бұрын

That's good practice for everyone

@quest4knowledge-xh3eu

4 ай бұрын

This stuff is tricky :)

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

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.

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

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

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

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

8 ай бұрын

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

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

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

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.

You explained the concept perfectly, thank you sir.

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

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

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

Awesome video and explanation, genuinely love your content!

Beautifully done. Thank you mate.

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

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

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

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

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.

Man this was incredibly helpful! Thank you so much!

@PrimeNewtons

8 ай бұрын

Glad it helped!

Great video! keep up the great work!

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.

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

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

@PrimeNewtons

4 ай бұрын

Haha. Thank you.

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

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.

legend 💪 keep it up man!!

earned a subscriber ,amazing teaching

Wonderful video. Thanks!

Excellent explanation. Subscribed.

thanks man, you helped me on my test

Best one, great mr, 💯💯🎉🥳

I love your channel so much.

@PrimeNewtons

3 ай бұрын

Thank you so much!

Amazing video ❤

Wow! I have no word to express you.

You're an absolute mathematical dude man!

@PrimeNewtons

10 ай бұрын

Lol. I see the pun.

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

Sir u are too much I learned a lot from you

saved my life thank you so much

Bro u are so good in maths ❤❤❤❤❤

Great video!

Super helpful!

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

Very good. Thanks 👍

🎉 yes oky please explain common shapes of absolut function

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

thank you for the help sir

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

Good stuff!

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

10 ай бұрын

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

great video

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

This guy is so good

@mabenj6954

7 ай бұрын

(no homo tho)

Thank you!!

can you do another epsilon delta proof

Thank you thank you thank you

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.

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

9 ай бұрын

I don't know

@Emine-ri7ex

9 ай бұрын

@@PrimeNewtons what it mean

@Noname61574

9 ай бұрын

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

Brilliant😁

AMAZING

helpful ty bro

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

Nice vid

From Angola 🇦🇴

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

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

7 ай бұрын

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

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

Se putea rezolva si prin explicitarea modulului?

👍👍👍

legend

Te amo.

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 😫

very good vi deo

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

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

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.

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

@MetalFaceDOOM3141

8 ай бұрын

I just realised it does not make a difference😂

thanks

@PrimeNewtons

7 ай бұрын

You're welcome!

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

10 ай бұрын

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

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

6 ай бұрын

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

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

خوش رجال

I'm waiting your response

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

Any jee aspirant here ? 😅

@niteshmayatwal

3 ай бұрын

Yes

I have never seen anyone make trivial math so complicated.

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