"Never stop learning, those who've stopped learning, have stopped living" 👏
@ummufaruq974 Жыл бұрын
I'm now understanding CALCULUS because of you,,,,Thank you so much.
@BelemaJumbo-kw1yb10 ай бұрын
Sir thank you very very very much, your video make complex things so easy to understand. God bless, and pls keep up the good work
@bitspieces2809 Жыл бұрын
His smile is like "this gon be good"😂
@jonathanchikwanda52697 ай бұрын
Lmao
@ImanuelNakwakiru5 күн бұрын
Your Genius 😮expencially the pascal triangle today have known the hidden trick thank you sir❤
@RowieFox3 ай бұрын
this was so great! I want to be a teacher and the way you go about teaching is so inspiring, thank you!
@thirugnanamn8358 Жыл бұрын
You are a beautiful math teacher. 👍
@PrimeNewtons Жыл бұрын
Thank you! 😃
@JijiJaja-fk5pt3 ай бұрын
Hey thank u for this video im from algeria , i like the way of ur teaching Dr like u simplify the information So thank you and good luck Dr Newtons
@brrrayday Жыл бұрын
I always love seeing the full derivation process, but I wanted to drop this pointer for those who may wish for a time saver: Since 1/x^3 is the same as x^-3 I did a little trick; I dropped the exponent in front of the x, and then subtracted one from the exponent's value, and put this -4 in the exponent position. It only works with a monomial though. f(x) = 1/(x^3) >>function to be derived 1/(x^3) >>rewrite as a monomial x^(-3) >>move exponent down -3x >>find value of new exponent -3 - 1 = -4 >>put it all together f'(x) = -3x^(-4) >>voila, your new derivative that little trick can save you a lot of time, but it only works in a very specific situation
@mausplunder5313 Жыл бұрын
thats how you would do it normaly. thats just the powerrule no?. the video is just a step back to look where our shortcuts like powerreduction multiply by power etc. stem from if we look at the defintion of a derivative by limits. you may have done some examples at the start of calc like this one but its actually useful for derivatives like sin(x).
@aammiranda6 ай бұрын
I know the math but the video is refreshing. Thank you
@SH-eo2py9 ай бұрын
U ARE AMAZING U SAVED ME THANK U SO MUCH
@moseschikusela9182 Жыл бұрын
Amazing lecturer
@StuartSimon Жыл бұрын
I used first principles to prove the power rule in the general case for positive integer exponents, and I’m hoping that this will help me do it in the general case for negative integer exponents.
@PrimeNewtons Жыл бұрын
math.mit.edu/~djk/18_01/chapter03/proof07.html
@PrimeNewtons Жыл бұрын
See if that helps your mission.
@MirriamMutale-j7i10 ай бұрын
You are the best 😘❣️
@shlomotenenberg Жыл бұрын
You are really good 👍
@DuaneDonaldson3 ай бұрын
I stop by your channel once in a while. If you watch some of the IVY LEAGUE university channels, you may notice that the smartest professors in the world keep their notes nearby if not in their hand. Teachers are not allowed to make mistakes like you have made several today. If you have experience in Japan and some other foreign countries, the teacher will keep their left finger on the negative sign to not forget to negate all calculations to the right. You will still be able to move your body out of the way so students can see your step by step solutions as you are working. Thank you for showing the Pythagorean Pyramid and the descending and ascending powers of the cubic expansion. Nice touch, cheers. THUMBS UP Without proofs, carry the power of 3 to the numerator and add 1 to the power at the denominator, and negate the answer, done. But I like your solution better.
@luketwomey50264 ай бұрын
great energy in the vid :)
@renesperb Жыл бұрын
Of course it is a good exercise the way you do it.But a more convenient way is to set y[x]= 1/x^3 .Then ,x^3*y[x]=1, and differentiation by the product rule givesy'[x]*x^3+3*y[x]*x^2 = 0 . A short calculation then leads to y'[x]=-3 y[x]/x = -3*1/x^4.
@ziggyharding5207 Жыл бұрын
Very helpful video, thank you!
@n0mvdslvnd584 Жыл бұрын
The Pascals Triangle was too fire!!
@bitspieces2809 Жыл бұрын
Never stop learning 👍
@JAMESYUN-e3t Жыл бұрын
Excellentl, Professor. By the way, let us get the chance to learn how math knowledge is limked with Python algorithms or R algorithms.
@SiloloSililo Жыл бұрын
what if we have 1/x-2 how do you solve that sir
@jonmoore8995 Жыл бұрын
Very good review!
@FantomJuma Жыл бұрын
This is really good
@roythoppilchacko835811 ай бұрын
Sir u may say like if we have fraction over fraction then division on top is multiplication in bottom that easy
@ClaraLukhere3 ай бұрын
Well understood
@panjak3233 ай бұрын
You know you can also do common denominator on the fraction subtraction
@abrahamicsantiago5653 Жыл бұрын
U good ❤
@dumebianthony-nwaokolo10727 ай бұрын
nice video
@said14121 Жыл бұрын
رائع ❤❤❤❤❤❤
@jan-willemreens9010 Жыл бұрын
... A good day to you Newton, I'm having your presentations as a part of my lunch time lately (lol); regarding this presentation it is indeed very smart/effective to multiply top and bottom by x^3(x + h)^3, thank you for this sort of advice. Busy day ahead today, so Newton also wishing you a pleasant day, and I'll be seeing you at the next challenging presentation ... Take care, Jan-W
@PrimeNewtons Жыл бұрын
Have a great week!
@ludzidzi Жыл бұрын
@@PrimeNewtons here in Namibia um enjoying your videos. Keep it up indeed u motivate me
@dionmaluleke-jz4jb8 ай бұрын
the final answer should be -3x to the power 2 innit ?
@MikeB35427 ай бұрын
Using the power rule, the exponent decreases by one...in this case from -3 to -4.
@LampuarJrFn-v1f2 ай бұрын
Much appreciated
@feddylacabra254 Жыл бұрын
Just WOW🤜🤛
@olayinkaadeleye3134 Жыл бұрын
Nice
@RicardoPerez-zc4ej11 ай бұрын
Very good. La imagen no es buena a veces.😢
@pjdrums68707 ай бұрын
thanks bro !
@ghoshfamily5755 Жыл бұрын
Really ❤
@sethdarlton Жыл бұрын
help me solve this, from the first principle of differentiation, solve f(X)=3
@vivekplays Жыл бұрын
I would believe that the absence of x would either make it undefined or 0. Due to the absence of x, there is nothing to substitute (x+h) into, and thus leaves you with either 0 or undefined. I'm not too sure which one.
@jumpman828211 ай бұрын
Yeah, it's a little intimidating because we end up with lim ℎ→0 [0 ∕ ℎ]. The key realization is that as ℎ _approaches_ 0, then 0 ∕ ℎ = 0, so we actually end up with lim ℎ→0 [0] = 0.