Derivative of 1/x^3 from first principles

In this video, I showed how to find the derivative of 1/x^3 from first principles. This process involves the use of basic binomial expansion using Pascal's Triangle

[@jensberling2341]

How I love Dr. Newton’s presentation. Thank you.

[@ummufaruq974]

I'm now understanding CALCULUS because of you,,,,Thank you so much.

[@bitspieces2809]

His smile is like "this gon be good"😂

[@jonathanchikwanda5269]

Lmao

[@aammiranda]

I know the math but the video is refreshing. Thank you

[@BelemaJumbo-kw1yb]

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

[@ziggyharding5207]

Very helpful video, thank you!

[@moseschikusela9182]

Amazing lecturer

[@StuartSimon]

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]

math.mit.edu/~djk/18_01/chapter03/proof07.html

[@PrimeNewtons]

See if that helps your mission.

[@SH-eo2py]

U ARE AMAZING U SAVED ME THANK U SO MUCH

[@jonmoore8995]

Very good review!

[@user-nj9ci3pv5e]

You are the best 😘❣️

[@shlomotenenberg]

You are really good 👍

[@user-sl4mg2ze7s]

This is really good

[@thirugnanamn8358]

You are a beautiful math teacher. 👍

[@PrimeNewtons]

Thank you! 😃

[@bitspieces2809]

Never stop learning 👍

[@n0mvdslvnd584]

The Pascals Triangle was too fire!!

[@feddylacabra254]

Just WOW🤜🤛

[@pjdrums6870]

thanks bro !

[@brrrayday]

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]

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).

[@dumebianthony-nwaokolo1072]

nice video

[@ghoshfamily5755]

Really ❤

[@user-il8mt2wz9t]

Excellentl, Professor. By the way, let us get the chance to learn how math knowledge is limked with Python algorithms or R algorithms.

[@said14121]

رائع ❤❤❤❤❤❤

[@roythoppilchacko8358]

Sir u may say like if we have fraction over fraction then division on top is multiplication in bottom that easy

[@abrahamicsantiago5653]

U good ❤

[@user-qt9un7ev6o]

what if we have 1/x-2 how do you solve that sir

[@jan-willemreens9010]

... 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]

Have a great week!

[@ludzidzi]

@@PrimeNewtons here in Namibia um enjoying your videos. Keep it up indeed u motivate me

[@renesperb]

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.

[@yemioladelee]

Thank you

[@PrimeNewtons]

You're welcome

[@olayinkaadeleye3134]

Nice

[@RicardoPerez-zc4ej]

Very good. La imagen no es buena a veces.😢

[@dionmaluleke-jz4jb]

the final answer should be -3x to the power 2 innit ?

[@MikeB3542]

Using the power rule, the exponent decreases by one...in this case from -3 to -4.

[@Red_Schuhart]

Witchcraft, clearly!

[@user-pt6nb6fc8g]

help me solve this, from the first principle of differentiation, solve f(X)=3

[@vivekplays]

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.

[@jumpman8282]

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.

[@zamokuhlelulama1800]

🙏🏽🤍

[@ghoshfamily5755]

Really ❤