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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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• if (vector v. f.) v = [x(t),y(t)] and f(x,y)=xy
is f(x(t), y(t)) = xy = f(v(t))?
• isn't the gradient a row vector? Otherwise how can you take the dot product
• It kind of depends. We can represent the gradient as a row or column vector, but like you mentioned we need to always be sure to line up our matrices in the right way as to make our matrix multiplication work out! :-)
• 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!
• 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.
• 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.
• Can you write all parametric functions as a vector function? How?
(1 vote)
• Yes!
Let's say that you have a parametric equation
` p(t) = {h(t)=2t, L(t)=3.5t+4, b(t)=7`
You'd define the vector function to be
`r(t) = <h(t), L(t), b(t)>`

You take the 1-nth functions in the parametric function and make them the corresponding 1-nth elements of the vector function.

I hope this helps!
(1 vote)
• 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)
• Why wouldn't one just substitute v into f, creating a scalar valued single-variable function, and then take the derivative normally?
(1 vote)
• 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).
(1 vote)