Multivariable chain rule intuition

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  • čas přidán 19. 05. 2016
  • Get a feel for what the multivariable is really saying, and how thinking about various "nudges" in space makes it intuitive.

Komentáře • 66

  • @melduck9701
    @melduck9701 Před 7 lety +182

    I should start paying you guys 9k a year instead of my uni

  • @effy1219
    @effy1219 Před 7 lety +52

    oh my god. this is the first time i saw ppl explaining chain rule this way. tho the image exits in my head i've never been able to draw it this clear!!!!!!!awesome!!!!!!!!!!!!

    • @abhishekravindra4008
      @abhishekravindra4008 Před 5 lety +1

      It is definitely a awesome way to look at partial derivatives chain rule.

  • @axelc6125
    @axelc6125 Před rokem +6

    I love randomly stumbling onto Grant videos when learning with Kahn academy. Its like seeing a popular celebrity you like at a coffee shop even though you both live in the same city. You know it not unusual that he's there, but its still a nice surprise.

  • @AC14xSkateOrDie
    @AC14xSkateOrDie Před 7 lety +19

    I've been confused by the notation for a while and this cleared everything up instantly!! This channel is the best, thanks for everything!

  • @BennettAustin7
    @BennettAustin7 Před 4 lety +25

    “You can just cancel out the dts” haha sounds like a physicist

  • @anonymoustraveller2254
    @anonymoustraveller2254 Před 6 lety +1

    Omg! I always had this in my mind but you put it on paper so correctly & its so amazing man ! Hats off

  • @iconbylvergara6188
    @iconbylvergara6188 Před 7 lety +7

    Thank you very much! You help us see the beauty in calculus :)

  • @alejrandom6592
    @alejrandom6592 Před 3 lety +3

    it's cool to see that the multivariable chain rule corresponds to the dot product between the gradient of f and the vector valued function derivative whose entries are the outputs of f

  • @mariadevina5594
    @mariadevina5594 Před 6 lety +2

    Thankyou so much all the time i am only remembering all the formula and i start to forget all. After knowing the core i can remember it for a along time thx!

  • @KVigulis11
    @KVigulis11 Před 7 lety +2

    That's the real way to explain it... Thanks!

  • @denielshumeiko6177
    @denielshumeiko6177 Před 4 lety +3

    The teacher from god... Really when people say that they don't understand math... the only reason they think so is because their teacher didn't manage to explain the whole logic behind some process... this was such an elegant way to establish the rule... it almost feels like you could have created this whole concept yourself and be a father of calculus in some sense

  • @FabianEason
    @FabianEason Před 6 lety +1

    Yes this is great. So much more intuitive.

  • @abhishekravindra4008
    @abhishekravindra4008 Před 5 lety +1

    You and only sal made sense of this topic to me

  • @user-iz2et4kj9y
    @user-iz2et4kj9y Před rokem

    Wow ! That completly changed my view!Thanks!

  • @futterkulcha
    @futterkulcha Před 17 dny

    Amazing explanation thank you!

  • @williamcole4689
    @williamcole4689 Před rokem

    thanks guys. This video just took the magic out of this concept.

  • @arturodeza3816
    @arturodeza3816 Před 7 lety +18

    This is also called the Total Derivative of a function 'f(t)' correct?

  • @iustinianconstantinescu5498

    I understood it intuively since the half of the video.

  • @chetanraina5649
    @chetanraina5649 Před 4 lety +7

    How can you just add the individual nudges? How do you know that they add linearly?

    • @That_One_Guy...
      @That_One_Guy... Před 4 lety +5

      Those variables (x, y, and the function f) are dependent of t (their value are determined by t), so every change in t would surely cause some change in x,y,f (f is also dependent of x and y so change in x and y are added for change in f). They add linearly because of the dependancy of t and the fact that derivative is a linear operator.
      Something is a linear operator if following is true : L(c*F) = c*L(F) and L(F+G) = L(F) + L(G) (L = linear operator, F and G = some function, c = some constant)

  • @observever7808
    @observever7808 Před 3 lety +1

    wait this is actually brilliant

  • @XahhaTheCrimson
    @XahhaTheCrimson Před 4 měsíci

    For those who wonder more strict representation of infinitesimal, watch 34th video of this list. He explained it more elaborately.

  • @andreasheirbrant2294
    @andreasheirbrant2294 Před 6 lety

    thank you very much!!

  • @cauchyschwarz3295
    @cauchyschwarz3295 Před 2 lety +2

    The cancelling out suggests that these two nudges are the same. Which they are not.

    • @vikraal6974
      @vikraal6974 Před rokem

      True, this is a poor analogy only helpful for memorization. But it's not the full picture

  • @Tshikumu
    @Tshikumu Před 4 lety

    WOW really nice thank you

  • @scholar-mj3om
    @scholar-mj3om Před 10 měsíci

    Marvellous💯

  • @ShBenEd
    @ShBenEd Před 2 lety

    I love you

  • @thesoul3461
    @thesoul3461 Před 4 lety

    awesome!!

  • @yashgutgutia604
    @yashgutgutia604 Před 3 lety

    Effing Legend mate

  • @neon_code
    @neon_code Před 3 lety

    Thanks!!

  • @kates-creates
    @kates-creates Před 10 měsíci +1

    3 blue 1 brown crossover !!!

  • @ShirimeCid
    @ShirimeCid Před 8 lety

    sweet!

  • @aburayhanmahadi7690
    @aburayhanmahadi7690 Před rokem

    Thank you sir

  • @XIC_OmPrakash
    @XIC_OmPrakash Před 3 měsíci

    lot more intutive than bookish definition of matrix multiplication

  • @felitham8685
    @felitham8685 Před 5 lety

    NICE

  • @SurajBorate-bx6hv
    @SurajBorate-bx6hv Před 2 lety

    Maybe a better way to explain this is that change in t is causing a nudge in f in the direction of v= dx/dt i + dy/dt j vector hence we find the direction derivative of f in the direction of v.

  • @sleepycorn349
    @sleepycorn349 Před 2 lety

    Thank you so much, 3b1b x khan academy!!

  • @aalaptube
    @aalaptube Před rokem

    Grant, I am still not clear why the plus sign in between. f(x, y) could be anything, i.e. x*y, x^y, or (x^2)*(3y^3)+2xy, literally whatever. Why would the + sign maintain?

  • @pedeeli177
    @pedeeli177 Před 4 lety +1

    hey is that the same guy as 3blue1brown, or does he just sound similar?

  • @user-jq4in7dm1v
    @user-jq4in7dm1v Před měsícem

    I don't understand, what happen if I just use df/dx and df/dy instead of the partial one?

  • @LaureanoLuna
    @LaureanoLuna Před 5 lety +3

    The point in need of proof is that the products of the partials dot dx/dt and dy/dt are to be ADDED. And no hint is given here.

    • @ykkrasaoz9748
      @ykkrasaoz9748 Před 5 lety

      Have to agree

    • @ElizaberthUndEugen
      @ElizaberthUndEugen Před 4 lety +1

      Indeed. He just snug the plus sign in without even mentioning it. And is the crucial part... bad.

    • @megapanka
      @megapanka Před 4 lety +4

      I believe it has to do with the fact they are independent variables (x does not depend on y) and as such we have we can add the partial effect of each one, giving us the total effect (variation of F(x(t),(y(t)) depends on x that depends on t and on y that depends on t. U add the individual effects and u get the total effect.

    • @crazyahhkmed
      @crazyahhkmed Před 4 lety +1

      I would think that was obvious though.

    • @LaureanoLuna
      @LaureanoLuna Před 2 lety

      @DnB and Psy Production Thnx.

  • @BeneathTheEquations
    @BeneathTheEquations Před 18 dny

    7:22 hey 3b1b!! you disappointed me there.

  • @KevinGodfreyVerpula
    @KevinGodfreyVerpula Před 29 dny

    I'm a little lost , doesn't the equation df = (df/dx) * (dx/dt) * dt + (df/dy) * (dy/dt) * dt , translate to df = 2df, when you cancel out the terms!

  • @rohitbhosle6521
    @rohitbhosle6521 Před 5 lety +1

    It sucks when each components of function i.e. f(x) f(y) and f(z) are function of x,y,z,t ,😫

  • @alejrandom6592
    @alejrandom6592 Před 3 lety

    5:38 did you just say constant? :c *sobbing function noises*

  • @HanBurritoz
    @HanBurritoz Před 6 lety +1

    Can you not just simply say that the derivative of the funtion is the same as the product of the directional derivate with the directional vector being the derivative caused in the change of t?

    • @agustinmiranda3989
      @agustinmiranda3989 Před 4 lety +1

      you mean dot product?

    • @joshuakaufman6787
      @joshuakaufman6787 Před 3 lety

      I believe that you can, as the multi variable chain rule can be expressed as a matrix multiplication, and in some cases, this represents itself as a dot product.

  • @adrienwillermain6604
    @adrienwillermain6604 Před 3 měsíci +1

    """lego batman meme"""

  • @sigurfeanaro3356
    @sigurfeanaro3356 Před 2 lety

    But dy/dt or dx/dt ARE fractions, they are infinitesimal fractions, they are the infinitesimal and indivisible part of a magnitude Δy/Δt and Δx/Δt.

  • @jeanphilippepuyravaud439
    @jeanphilippepuyravaud439 Před 3 lety +1

    The trouble with your videos is that it becomes impossible to look at other maths videos...

  • @knowit3887
    @knowit3887 Před 3 lety

    700th like

  • @carbon273
    @carbon273 Před 4 lety

    Hmm I think Patrick did it better.🤔