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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. Created by Grant Sanderson.

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  • primosaur ultimate style avatar for user uncinoO
    if (vector v. f.) v = [x(t),y(t)] and f(x,y)=xy
    is f(x(t), y(t)) = xy = f(v(t))?
    (5 votes)
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  • aqualine tree style avatar for user Jordan Purser
    At how come when we take the derivative of the vector valued function on the left side we get a vector of the respective derivatives of the variables, but when we take the derivative of the parametric equation on the right side we get a dot product of the gradient with the vector of the derivatives of the variables? I though the vector valued function and the parametric equation were just the same thing in different notation? I'm missing something somewhere!
    (3 votes)
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    • leaf green style avatar for user prasook
      Hi Jordan, as you are saying vector valued function v(t) = x(t) i + y(t) j, and parametric equations x(t), y(t) which define xy plane, both are same because they both defining position of a point on xy plane.
      But at right hand side, there is a new introduced function f, inputs are the xy plane points, so in the video grant first take parametric equation of x and y coordinates and write the chain rule equation, then transform it as input of into vector valued functions.
      (2 votes)
  • duskpin seed style avatar for user Kesleta3
    Why wouldn't one just substitute v into f, creating a scalar valued single-variable function, and then take the derivative normally?
    (2 votes)
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    • old spice man green style avatar for user Elijah Daniels
      I asked this exact same question in my Calc III class. The answer is yes, of course you could and make your life easier, but that doesn't show you the importance of learning how to do the multivariable chain rule. It's important to learn because you aren't always given the definitions of the other functions. Sometimes you are given one multivariable function and you are asked to find the partial of one variable with respect to another, and this is a case where you don't simply have a function to plug in to another (a fun example, the partial of pressure with respect to volume in the van der Waals equation).
      (2 votes)
  • blobby green style avatar for user 정성은
    Can you write all parametric functions as a vector function? How?
    (1 vote)
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  • blobby green style avatar for user ccbb3899
    isn't the gradient a row vector? Otherwise how can you take the dot product
    (1 vote)
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  • aqualine ultimate style avatar for user John He
    I think Grand has to be careful with putting a vector into the gradient, since it's quit confusing to take partial derivative of a vector.
    (1 vote)
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  • boggle blue style avatar for user Bryan
    Is the the gradient of f evaluated with v(t) (where v is a vector-valued function) the same as gradient of f evaluated with a point, where the point's x, y, z, etc coordinates are just the corresponding components of v(t)?
    (1 vote)
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  • female robot grace style avatar for user I. Khurram
    At instead of writing:
    Can you write:
    ∇v(t) ?
    (0 votes)
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Video transcript

- [Voiceover] So in the last couple of videos, I talked about the multi-variable chain rule, which I have up here, and if you haven't seen those go take a look. Here and I want to write it out in vector notation, and this helps us generalize it a little bit when the intermediary space is a little bit higher dimensional. So, instead of writing X of T and Y of T as separate functions, and just trying to emphasize "oh they have the same input space, and whatever X takes in that's the same number Y takes in." It's better and a little bit cleaner if we say there's a vector valued function that takes in a single number "T," then it outputs some kind of vector. In this case you could say the components of V are X of T and Y of T, and that's fine. But I want to talk about what this looks like if we start writing everything in vector notation, and just since we see DX/DT here and DY/DT here, you might start thinking, "oh we should take the derivative of that vector valued function." The derivative of V, with respect to T, and when we compute this it's nothing more than taking the derivatives of each component. So in this case, the derivative of X, so you'd write DX/DT, and the derivative of Y, DY/DT. This is the vector value derivative. And now you might start to notice something here. Okay so we've got one of those components multiplied by a certain value and another component multiplied by a certain value, you might recognize this as a dot product. This would be the dot product between the vector that contains the derivatives, the partial derivatives, partial of F with respect to Y, partial of F with respect to X, oh, whoops, don't know why I wrote it that way, but up here that's with respect to X, and then here to Y. So this whole thing, we're taking the dot product with the vector that contains ordinary derivative DX/DT and ordinary derivative DY/DT. And of course both of these are special vectors, they're not just random, the left one, that's the gradient of F, and the right vector here that's what we just wrote that's the derivative of V with respect to T, just for being quick I'm gonna write that as V prime of T. That's saying completely the same thing as VDVT, and this right here is another way to write the multi-variable chain rule, and maybe if you were being a little bit more exact you would emphasize that when you take the gradient of F the thing that you input into it is the output of that vector valued function, you know you're throwing in X of T and Y of T, so you might emphasize that you take in that as an input, and then you multiply it by the derivative, the vector valued derivative of V of T. And when I say multiply, I mean dot product, right, these are vectors and you're taking the dot product, it should seem very familiar to, you know, the single-variable chain rule. And just to remind us I'll throw it up here, if you take the derivative of composition of two single-variable functions F of G, you take the derivative of the outside F prime, and throw in G, throw in what was the interior function, and you multiply it by the derivative of that interior function, G prime of T. And this is super helpful in single-variable calculus for computing a lot of derivatives, and over here it has a very similar form right? The gradient which really serves the function of the true extension of the derivative for multi-variable functions for scalar valued multi-variable functions at least. You take that derivative and throw in the inner function, which just happens to be a vector valued function. You throw it in there, and then you multiply it by the derivative of that, but multiplying vectors in this context means taking the dot product of the two, and this could mean if you have a function with a whole bunch of different variables, let's say you have some F of X, or not F of X, F of X1 and X2 and it takes in a whole bunch of variables that it goes out to X100. And then what you throw into it is the vector value function that's vector valued, takes in a single variable, and in order to be able to compose them it's gonna have a whole bunch of intermediary functions, and you can write it as X1, X2, X3, all the way up to X100 , and these are all functions at this point. These are component functions of your vector valued V. This expression still makes sense, right? You can still take the gradient of F, it's gonna have 100 components, you can plug in any vector, any set of 100 different numbers, and in particular the output of a vector valued function with 100 different components is gonna work, and then you take the dot product with the derivative of this. That's the more general version of the multi-variable chain rule, and then the cool way about writing it like this, you can interpret it in terms of the directional derivative, and I think I'll do that in the next video, so, that's a certain way to interpret this with a directional derivative.