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Find (a) $f_{x x}(x, y),$ (b) $f_{y y}(x, y),$ (c) $f_{x y}(x, y),$ and $f_{y x}(x, y)$ for the function defined in the given Exercises.Exercise 1

(a) $4+24 x y^{4}$(b) $6+48 x^{3} y^{2}$(c) $48 x^{2} y^{3}$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 2

Partial Derivatives

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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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So if you haven't watched the video for exercise one, I would go to that because we walk through how we get these first partials here because we'll need these partials in order for us to find the second partials. Eso Let's see, what order did they want to do? So right, So first thing was to find the second derivative with respect to X. Um, so let's come over here and do that. So if we come over here and take the partial of this with respect to X, I'll give us the second partial. And so now remember, we assume that this why here is a constant, um So when we go through and take the derivatives of this, we'll just treat as it's a constant, just like this nine. So four X would just give us four, and then we have plus at all factor this y to the fourth out. So B 12 y to the fourth del by Dell X of X squared and in the derivative of nine is just going to be zero. Then we could come over here and take the derivative of X squared, which should just be two x that we could go ahead and multiply these together so it be four plus 24 x y to the fourth so that this is going to be our second partial derivative. All right, um and so this was a now for B. We come over here and then take the second partial of this with respect. Why? So again, remember, we assume this X is going to be a constant. Um, so we have, uh why, Why so six y would just be six, and then we go ahead back to that exile so would be plus X cubed del by del y of y cubed. And then the derivative of two is just going to be zero. And then this here is going to be three y squared that we could go ahead and multiply those together. So it be six plus, um 48 execute y squared. So this is our second partial. Their respective why? And then it wants us to find the partial with respect to X. Then why and why the next? And so this should be the same value or the same equation that we get out. But let's just double check to see if that is true. So let me pick this up here and then scoop this down. I don't do the same thing over here, right? So instead of having the same one, we're going to take the opposites over here. I'm going to do Del by Dell. Why this time? Now, let me just get rid of these arrows. So we're going to assume these excess or constants. So when I take the derivative that there should be zero so f y x is going to be zero, Um, there will be plus 12 x square Del Beidle. Why of why? To the fourth and then the derivative nine. It's just gonna be zero. And then this would be for why cute and so that multiplying everything together we get 48 squared. Why acute? So this is our partial first with X that why and then if we come over here and do the same thing, but we do it with X this time. So we're going to treat why, as our constant. So we have f x y would be that would be zero plus 16 y cube del by Dell X of execute and that derivative to zero. If we take this here, that should give us three acts squared and so multiplying that altogether, you see, how are mixed? Partial ends up being the same thing. So it really didn't matter in which way we took this partial. Um, so it's normally just whichever one looks easier to take the partial off. Um, yeah, you could see that we get the same thing regardless.

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