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!
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
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.
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 :)
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.
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.
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.
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
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.
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
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?
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 😫
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.
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.
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
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!
That's good practice for everyone
This stuff is tricky :)
Your videos definitely deserve much more attention, I feel lucky that youtube reccomened me your 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
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
Omg 😭 i had no idea you could write absolute value like that, thanks so much, now I can really handle these problems
came looking for an answer to my question and the explanation answered before the video started. thank you!
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 :)
Thank you for the feedback. Hope to keep improving every day
Never ever get demotivated brother remember a person from india will always be your no 1 supporter❤❤❤ brother injust subscribed❤
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.
Great explanation . With your help you help a lot of people struggling with math...no doubt. Cheers from Belgium
I like the energy of this video. Very good explanation. Thank you
You explained the concept perfectly, thank you sir.
Awesome video and explanation, genuinely love your content!
thank you for your effort, it is so clear and helpful!
you and JG are the best i've ever seen on youtube
Beautifully done. Thank you mate.
Great video! keep up the great work!
The key is the deffinition of the absolute value. A fine work, indeed!
legend 💪 keep it up man!!
Man this was incredibly helpful! Thank you so much!
Glad it helped!
awesome class man! For a moment I felt like Kamaru Usman from UFC was teaching me a lesson xD. Joke aside, neat explanation.
Best one, great mr, 💯💯🎉🥳
He got it in him, a way of making maths looking too easy, fun and enjoyable.
Thank you Sir for the great explanation. Never stop learning . All the best to you
thanks man, you helped me on my test
earned a subscriber ,amazing teaching
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.
Excellent explanation. Subscribed.
Wow! I have no word to express you.
Dear your personality and also your acting skills are amazing you should be in Hollywood movies.
Haha. Thank you.
Wonderful video. Thanks!
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
Amazing video ❤
I love your channel so much.
Thank you so much!
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.
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
The absolute truth is Prime Newtons is number one! 🎉😊
Sir u are too much
I learned a lot from you
saved my life thank you so much
Super helpful!
You're an absolute mathematical dude man!
Lol. I see the pun.
Great video!
Bro u are so good in maths ❤❤❤❤❤
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.
Very good. Thanks 👍
Good stuff!
This guy is so good
(no homo tho)
thank you for the help sir
great video
very good explanation
Thank you!!
is it the same for || x ||, norms of a vector?
🎉 yes oky please explain common shapes of absolut function
👍👍👍
Thank you thank you thank you
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
helpful ty bro
Brilliant😁
AMAZING
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
Aww. I'm sorry. You are better equipped for next time.
When you learn something you basically already knew, but never thought about.
I feel so terribly _inside_ the box right now.
lifesaver
legend
From Angola 🇦🇴
Nice vid
can you do another epsilon delta proof
Wouldn’t it be easier to just view the absolute value function as a piecewise function?
Te amo.
im sorry,but ive been avoiding this, yhooo you r so handsome
very good vi
deo
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?
I don't know
@@PrimeNewtons what it mean
i assume you can implement quotient rule and chain rule to get an answer
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
For all real x, the square root of x² is +x or -x. Which is |x|
Se putea rezolva si prin explicitarea modulului?
'u' substitution seems a bit unnecessary since the chain rule works fine and it is often introduced earlier in the class. Fine proof anyway
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 😫
thanks
You're welcome!
I thought the dx of abs(x) is the sign function and therefore abs(x)/x?
I just realised it does not make a difference😂
Sir please I need videos on partial differentiation please sir help me...... Multivariable calculus
خوش رجال
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.
Simplifying a function changes it. f(x)=sqrt(x^2) and f(x)=x^1 is not the same function.
Sir can you once explain about second order differentiation
Search for videos in my channel
@@PrimeNewtons okay sir, thank you 😊
Sir can you do a video on hw to evaluate square root limits eg limx~0(√x^2+9)-(2x^2+3/x^2))
Search in my channel. I already have many videos on that
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.
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'm waiting your response
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.
I don't follow what is happening here, I kept replaying the video but nothing to understand I just gave up.
x/|x| = sign(x)
= -1 when x < 0,
1 when x >= 0
Any jee aspirant here ? 😅
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.