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Find (a) $f_{x}(x, y),$ (b) $f_{y}(x, y)$ at the indicated point.$$f(x, y)=3 x^{4}-2 y^{3}-7 x^{2} y^{5}+8 x^{2}-3 y-5(2,1)$$

(a) 100(b) -149

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 2

Partial Derivatives

Campbell University

Harvey Mudd College

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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Yeah. So this question, in some sense, is a continuation of number two. So if you haven't watched number two Oh, are done problem too. I would go ahead and do those because we've already actually solved for these in that problem there. So I just kind of skipped the work and kind of say what we got for our result already. Um, so these are the partial derivatives that we got. And then if we want to find the partial at this point to one, all we need to do is plug in. So this is X, and this is our Why actually, let me go ahead and scoot. He's down a little bit. So when we do half sub x of 21 silver is gonna plug the numbers and so would be 12. And then too cute, which is eight minus 14 times two and then one to the fifth, which is just one of one. And then 16 times, too. Uh huh. Um, should let me sit down again. Then we would just add all of these eso 96 minus 28 plus 32 then that would be four plus 96. So 100. So are partial derivative at +21 is going to be 100 and then our partial with respect. Why? Same thing was gonna plug into one. Um, so you know, f sub y of 21 and then negative six. And then one square is just one minus 35. Um, and then two squared is going to be four times one to the fourth one. Nice three on, then. This should be so just negative. Six minus 1 40 minus three. So that would be negative. 1 49. So then are partial with respect, toe. Why at the point, um, to one is gonna be negative. 149. So again, uh, if you haven't watched number two or attempted to do number two you I would go and watch that video if you were curious as to how we actually got, um, this equation here and then this equation here. Just refer back toe problem number two

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