Vector form of the multivariable chain rule

The multivariable chain rule is more often expressed in terms of the gradient and a vector-valued derivative. This makes it look very analogous to the single-variable chain rule.

[@user-or7ji5hv8y]

Wow, I wish I had this resource when I was struggling through calculus.

[@ManojKumar-cj7oj]

The best part it is free indeed

[@lunacui461]

it enlight me a lot!!! especially when you lead to form of the dot product and the general chain rule...Thank you very much!

[@philandthai]

r(t) to denote a position vector would probably be better than v(t) which suggests velocity, granting that it is just a letter name.

[@amj5915]

Ah yes, Grant Sanderson. The goated one

[@scholar-mj3om]

Marvellous💯

[@orkhanruth]

Isn't this guy from "3Blue1Brown" ?

[@atomslaughter5892]

I think he worked for Khan for awhile before doing his own thing. Although I'm responding two years later so by now you probably either know that or do not care lol

[@pooper2831]

@@atomslaughter5892 No, he left youtube after not getting reply for 2 years. Sorry for responding to your comment for another two years.

[@tomdwyer8809]

@@pooper2831 hey man, I’m not sure what you mean. Not sure if I’ll get a response since your comment was from 2 years ago

[@gytcy7592]

@@tomdwyer8809man you’re quick, I was going to continue this thread. Guess I have to wait another 2 years then.

[@anonymous9217w2]

@@gytcy7592 reminder 3 months over.

[@alexdellatorre9996]

I don't think he'd talked about gradients before, had he?

[@sgut1947]

Yes, for example in videos 19--25 in the multivariable calculus playlist.

[@urieldaboamorte]

I love this man but why he always gotta multiply two column vectors 😭

[@98danielray]

because he is right. thats how inner products work. there is a distinction between matrix multiplication and inner products, hence why the dot appears and why x•y is tipically denoted x^Ty when in euclidean space

[@krussell3612]

@@98danielray Don't you need a particular sign to reference Hadamard product, the binary operation that he uses in the video rather than the product of vectors?

[@solsticetwo3476]

Gradient should be a row vector. Also, it is presented as the directional derivative of f(r) following v'(t). But, what is not clear is how to evaluate a vector v(t)=(x,y) on another vector (grad f).

[@GamerTheTurtle]

horribly pedantic, the confusion of seeing grad(f) as a column matrix is negligible to any reader who actually knows what they're reading

[@Labroidas]

@@GamerTheTurtle This video is for people who are learning, so confusing elements like these should be weeded out. If you do a product of two column vectors, you get a vector, not a scalar, so this is actually a big problem. Consistency is the most important thing in teaching and learning.

[@98danielray]

the gradient is not a row vector
df is

[@98danielray]

@@Labroidas 1- this is not a significant distinction
2- the gradient is NOT a row vector

[@98danielray]

@@Labroidas also, there is no standard product of column vectors. you are thinking of the inner product, which IS with twocolumn vectors

[@Bunk_Moreland]

you've written V(t), but shouldn't it be V(x,y) ?

[@zairaner1489]

V is a fucntion which depends on t and not on x and y, so v(t) is correct. f depends on x and y, so f(x,y)

[@sigurfeanaro3356]

@@zairaner1489 v(t) is a vector function describing vectors in which the functions x(t) and y(t) compute the tails and heads points of the vectors of that vector function «It's just a way of describing the same thing in a fancier way»? Or am I wrong? Mathematicians like to complicate things.

[@cauchyschwarz3295]

I really love 3b1b but its kinda dirty here. You're writing the dot product between two operators as if they were vectors in R2. This helps only in how to use this. Not to understand it.