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Find the partial derivatives with respect to (a) $x$ and (b) $y$.$$f(x, y)=2 y^{2}\left(2 x^{3}-3 y^{4}\right)^{5}$$

(a) $60 x^{2} y^{2}\left(2 x^{3}-3 y^{4}\right)^{4}$(b) $4 y\left(2 x^{3}-3 y^{4}\right)^{4}\left(2 x^{3}-33 y^{4}\right)$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 2

Partial Derivatives

Missouri State University

Campbell University

University of Nottingham

Idaho State University

Lectures

12:15

In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

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Find the partial derivativ…

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05:03

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Find all the second partia…

01:54

Find the indicated partial…

03:21

00:43

Find the first partial der…

So remember, whenever we're going to take a partial derivative of something, So if we're going to do the partial derivative of this with respect, X, what we assume is all other variables or actually, Constance. So this y squared here in this y to the fourth. We're going to treat both of those just that they were a constant. So let's first go it in right out. This is so a lot of times you'll see this written is Dell F of Dell X, um, or which I think the more common way to rise F Sub X like that now, since two Y squared is a constant with respect X, we can pull that out. So it's two y squared del by Dell X of to execute minus three y to the fourth, all raised to the fifth. And then we could go ahead and take the derivative of this using channel. So we have two squared move the five out front to execute minus three wide to the fourth and then subtract, um five or subtract one from fives. We get four, and then remember, we take the derivative of the inside and multiplied by that, um so that would be six x squared. And now why is treated as a constant? So why to the four still constant multiplied by constant still constant. So that would just be zero. Now, if we go ahead and multiply all this together, um, 10 times six is 60 and then x squared y squared to execute minus three y to the fourth, all raised to the fourth. And so this is going to be our partial with respect to X and unfortunately for us, are partial with respect. Why will not be as easy? Because now, if we were to change this to be partial perspective, Why, um, the only thing that's really going to be a constant is this x here now? So we have two functions that depend on why being multiplied together. So we would need to use product rule. So let's go on to write that up. So we have dlf by del y Or does he? Quito s a boy? Eso we'd have to Why Square del Baidoa y of to execute minus three wide Fourth all raised fifth and then plus where we have these flips of to execute minus three wide to the fourth all race the fifth del by del y of two y squared eso I'll just do this one over here first, because that's the easier one to do. So we just use power rules out before why? Um and now over here will need to do channel just like we did before s it would be five times to execute minus three y to the fourth, all to the fourth. And then remember, when we take the derivative of the inside here, we treat X as a constant now. So it be zero, then minus 12 y cute. And now we can go ahead and multiply everything together. Actually, before we do that, I'm just gonna go ahead and pull out. Um, these here, er least four copies of it. So I don't have to write that out a bunch of times on Also, pull out a looks like I can pull out a four years. If I pull this far out and a four out from here, that will work to say I will have four times to execute minus three y to the fourth, all to the fourth. And then after that, eso if we pull a four out. They'll just be three. So it would be two times, five times three, which be 30 on. Then we would have y squared times three y, which would be y to the fifth. I think that's all we have for the first one. And then over here, if we pull out that form that for just goes away, and then we would just have one set of this times. Why that be? Why? Times to execute minus three y to the fourth. And actually, let's just go ahead and pull out this. Why here? So that would make that a four. And then why just comes out front like that? Um, And then when you just go ahead and add these white, actually, this should be a negative because we had the negative up top here. Almost forgot that. Yeah. Yeah. Now we can go ahead and combine these white to the four terms, and then the to cube just kind of stays, um, so we would end up with so for why? To execute mines three y to the fourth all race the fourth, and then negative 33 y to the fourth plus two x cute. So then this here will be our partial with respect to why

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