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
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).
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.
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.
... 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
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.
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.
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.
"Never stop learning, those who've stopped learning, have stopped living" 👏
How I love Dr. Newton’s presentation. Thank you.
I'm now understanding CALCULUS because of you,,,,Thank you so much.
His smile is like "this gon be good"😂
Lmao
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
this was so great! I want to be a teacher and the way you go about teaching is so inspiring, thank you!
You are a beautiful math teacher. 👍
Thank you! 😃
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
Amazing lecturer
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
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).
I know the math but the video is refreshing. Thank you
U ARE AMAZING
U SAVED ME
THANK U SO MUCH
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.
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.
math.mit.edu/~djk/18_01/chapter03/proof07.html
See if that helps your mission.
You are really good 👍
You are the best 😘❣️
great energy in the vid :)
The Pascals Triangle was too fire!!
Very helpful video, thank you!
Very good review!
Excellentl, Professor.
By the way, let us get the chance to learn how math knowledge is limked with Python algorithms or R algorithms.
This is really good
what if we have 1/x-2 how do you solve that sir
Never stop learning 👍
... 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
Have a great week!
@@PrimeNewtons here in Namibia um enjoying your videos. Keep it up indeed u motivate me
Well understood
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.
nice video
رائع ❤❤❤❤❤❤
Sir u may say like if we have fraction over fraction then division on top is multiplication in bottom that easy
You know you can also do common denominator on the fraction subtraction
U good ❤
Much appreciated
the final answer should be -3x to the power 2 innit ?
Using the power rule, the exponent decreases by one...in this case from -3 to -4.
Nice
Just WOW🤜🤛
Very good.
La imagen no es buena a veces.😢
Really ❤
thanks bro !
Thank you
You're welcome
help me solve this, from the first principle of differentiation, solve
f(X)=3
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.
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.
Witchcraft, clearly!
🙏🏽🤍
Really ❤