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Multivariable calculus
Course: Multivariable calculus > Unit 2
Lesson 7: Partial derivatives of vector-valued functions- Computing the partial derivative of a vector-valued function
- Visual parametric surfaces
- Partial derivative of a parametric surface, part 1
- Partial derivative of a parametric surface, part 2
- Partial derivatives of vector valued functions
- Partial derivatives of vector fields
- Partial derivatives of vector fields, component by component
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Computing the partial derivative of a vector-valued function
When a function has a multidimensional input, and a multidimensional output, you can take its partial derivative by computing the partial derivative of each component in the output. Created by Grant Sanderson.
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Video transcript
- [Voiceover] Hello, everyone. So what I'd like to do
here and in the following few videos is talk about how you take the partial derivative of
vector valued functions. So the kind of thing I have in mind will be a function with
a multi-variable input, so this specific example have
a two variable input, p and s. You could think of that
as a two-dimensional space as the input or just two separate numbers. And its output will be three-dimensional. The first component, p
squared minus s-squared. The y component will be s times t. And that z component
will be t times s-squared minus s times t-squared,
minus s times t-squared. And the way that you
compute a partial derivative of a guy like this, is actually
relatively straight-forward. If you're to just guess
what it might mean, you'll probably guess right. It will look like partial
of v with respect to one of its input variables,
and I'll choose t with respect to t. And you just do it component-wise, which means you look at each component and you with a partial derivative to that 'cause each component is just a normal scaler valued function. So you go up to the top one and you say t-squared looks like a variable, as far t is concerned,
and this derivative is 2t. But s-squared looks like a constant, so its derivative is zero. S times t, when s looks like a constant and when t looks like a variable, has a derivative of s. Then t times s-squared,
when t's the variable and s is the constant, just
looks like that constant, which is s-squared
minus s times t-squared. So now a derivative of t-squared is 2t and that constant s stays in. So that two times s times t. And that's how you compute it, probably relatively straightforward. The way you do it with
respect to s is very similar, but where this gets fun
and where this gets cool is how you interpret the
partial derivative, right, how you interpret this
value that we just found. And what that means
depends a lot on how you actually visualize the function. So what I'll go ahead
and do in the next video and in the next few ones, is talk about visualizing this function. It'll be as a parametric surface and three-dimensional space. That's why I got my
grapher program out here and I think you'll find
there's actually a very satisfying understanding
of what this value means.