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Find the partial derivatives with respect to (a) $x$ and (b) $y$.$$f(x, y)=8 x^{2} e^{-2 y^{2}}$$

(a) $16 x e^{-2 y^{2}}$(b) $-32 x^{2} y e^{-2 y^{2}}$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 2

Partial Derivatives

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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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Find the first partial der…

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So in order for us to take the partial derivatives of these, remember, what we're going to do is we assume that all of the other variables are going to be Constance. So if we want Thio first, find the partial of this perspective X. So we apply the Dell by Dell X on both sides. So on the left, we have f sub X to represent Mr Partial with respect X. Then over here, well, we're going to be treating all these wise as constant. So this e to the negative two y squared is a constant. So we would just kind of pull that out and then just take the derivative of X squared. So let me just write that out. So the eight times e to the negative two y squared del by Dell X of X squared. And now we can take the derivative of this just as we would end just single variable calculus, which is going to give us two acts. Now we can multiply everything together on that will give us 16 x e to the negative two y squared. And so this is going to be our partial with respect to X now, in order for us to get our personal respect toe Why, instead of assuming that X is our variable And why is r constant? We're going to flip that around. So when you write that over here So Dell Beidle Why so now this eight x squared, we assume, is our constant. So we'd have f sub y is equal to so eight x squared del by a y of e to the negative two y squared And then to take the derivative of this will have to use changeable. So it be eight x squared And now just be e to the negative two y squared. But then we take the derivative of the inside s o negative two y squared would be negative. Or why that we could go ahead and multiply stuff together. So negative 32 x squared. Why e to the negative e to the negative tube, Why squared? And now this here is our personal with respect to why and so again, remember when we're going through and we're taking these partial derivatives, whichever partial we're taking, we assume everything else is a constant. And then just apply all of the rules we've learned so far from, um, like single variable calculus

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