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Find all the second partial derivatives_ f(x,Y) = xy - 3x5y2fxx(x, Y) = 12x2y _ 60x3,y2fxylx, y)4x3 30xfyx(x, Y)4x3 30xfyx(x, Y)6x5Need Help?ReedliSubmit Answer[~/2...

Question

Find all the second partial derivatives_ f(x,Y) = xy - 3x5y2fxx(x, Y) = 12x2y _ 60x3,y2fxylx, y)4x3 30xfyx(x, Y)4x3 30xfyx(x, Y)6x5Need Help?ReedliSubmit Answer[~/2 Points]DETAILSSCALCET8 14.3.063_Find the indicated partial derivative(s). f(x,Y) = xy5 (xxx' Txyxfxxx(x, Y)fxyx(x, Y)

Find all the second partial derivatives_ f(x,Y) = xy - 3x5y2 fxx(x, Y) = 12x2y _ 60x3,y2 fxylx, y) 4x3 30x fyx(x, Y) 4x3 30x fyx(x, Y) 6x5 Need Help? Reedli Submit Answer [~/2 Points] DETAILS SCALCET8 14.3.063_ Find the indicated partial derivative(s). f(x,Y) = xy5 (xxx' Txyx fxxx(x, Y) fxyx(x, Y)



Answers

Find all the second partial derivatives.
$$
z=\frac{y}{2 x+3 y}
$$

So we have B of X y is equal to x times alli over x minus y and we won't find all second partial derivatives. Eso In order to do that, we first need to find V X and B So let's start with the ex. So we feel we have a fraction here, so we need to use the closure rule. So we take the bottom times the derivative of the tops with respect to X that gives us why, minus the top times, the derivative of the bottom. With respect, X gives us one and then we square what's on the bottom. So this gives us a first derivative with respect to X uh, x y minus y squared minus X y over X minus y squared. Then we can cancel our ex wives so we get negative y squared over X minus y quantities. And then let's find the, uh, partial derivative with respect. Why? So the sound we're treating X is a constant, and we're going to use the product rule against. So we have the bottom times, the derivative of the top with the respect Thio. Why giving us X minus the top times the derivative of the bottom with respect. Why gives us negative one and square what's on the bottom? So we have the same denominator. So we see we have X squared a man with minus x y plus it's why so the ex wise cancel again over X minus y eyes. So now that we have these, we confined our partial derivatives. So let's find V x X. It's so we're taking the derivative with respect to X uh, negative y squared over X minus y squared. So let's use the closure. We have the bottom times. The derivative of the tops with respect to X is your zero minus the top times the derivative of the bottom with respect to X. So we have to has the quantity x minus y and then using the chain rule, the derivative of the inside with respect to X just gives us one, and then you square with on the bottom. So instead of squared, we have X minus y to the fourth. This is zero's. We can cancel that out, and we could cancel this X minus wire with one of these ones leaving us with to y squared over X minus y puked. So that's one of our second partial derivatives. Um, next let's do speed. Why? Why? So we're taking the road of with respect. Why, uh, X squared over X minus y squared. So using the question rule again, we have the bottom times the druid over the top, which is zero minus the top times, the derivative of the bottom, which is again, too times X minus y. But now we take the derivative what their sector while on the inside, giving us a negative one. And then we square what's on the bottom. So we have X minus y to the fourth again. First term zero negative cancels with the negative, um, X minus y cancels with one of those giving us two x squared over x minus y keep. And we know that the cross derivatives be X y is equal to be why ex So we could just pick one of those defined. So let's do the derivative with respect to X of B y, which was X squared over X minus y squared. So if we take this derivative with respect to X, but use the closure rule so we have the bottom times the derivative of the top With respect X, We're just two x minus a cup times the derivative of the bottom with respect to X. So we have two X minus y times that a relative of the inside with respect acts, which just gives us one over the bottom squared, which is X minus y to the fourth's. Ah so simplifying. We can factor out on X minus y so we can cancel one out from each term. So two becomes a one and four becomes a three. So we end up with two x times. That's minus why minus two X squared over X minus y cute. And then we see we'll get her to X squared here so we can cancel the two x squared. So the top becomes two x times negative. Why gives us negative two x? Why? So we end up with negative to X y over X minus Y keeps for our cross derivatives

So in order for us to find the partials with respect to both x and y of this, nor will we end up doing is we treat all of the other variables as if they're constants, and then we just take the derivative like we would in the one variable case. So let's come over here and first do the partial with respect. X o Dell by Dell X It's on the left. We would normally write. This adds, um f sub x. So the partial with respect to X um And then we're going to treat this Why? And this Why? Just as if they were Constance. So at first we could just go ahead and pull out, too. Why, from this would be to i del by Dell x of x times e to the three x. Why, Yeah, And now we have a function that depends on X multiplied by a function that also depends on X. So we use product rule. So would be to why, uh, times so it be Extell by Dell X of E to the three x y, and then plus E to the three x y del by Dell X Now to take the derivative of E 23 x y. So we first take the derivative of the outside because we need to use change rules. So that doesn't change to those still be e 23 x y But then, remember, change will just take the derivative of the inside and we're going to treat this. Why is a constant So it's like three times Why both? Constant. So we just take the derivative of X essentially, which will give us three. Why? Andan over here taking the derivative of X just gives us one. So let's go ahead and raped this out. So you have to y times it looks like three x y AII to the three x y plus e to the three x y And then we could go ahead and factor out that three to the ex wife because there is a common factor. I mean, not really needed, but just to clean it up a little bit. So we have to i iii to the three x y And then times, um three x Y plus one. Yeah. Uh huh. And so this is our partial with respect to X right? Next, we can go ahead and take partial with respect to why eso them come appear and pull this down. And so now, instead of treating these wise is Constance, we treat the wise is our variables that we're going to treat the exes, as are constants. So we do, Del by Dell X off this. So you're not dealt by dogs? Where did that Dell? By don't Why of that? So we have f sub y or the partial of f with respect to buy is able to remember this x and this expatriates constants. So let's just pull it out just like we did before the two x tell by del y of y AII to the three x times wife. And again we have a function that depends on why multiply function. That depends on why. So we use product raw. So this will be two x times. So why Dell? Beidle Why, uh v 23 x y and then plus E to the three x y times. Uh del Beidle, Why of why? And again this derivative here is going to be essentially the same thing over and just have an X there instead. Um, so we have e 23 x y and then we take the road of the inside Do the chain rule eso the X now is a constant so we just take the derivative wise would be three x And then over here, del by del Y y is just one And now we go ahead, multiply everything together as we have two X um Times three x y e to the three x y plus e +23 ex wife that we could go ahead and factor out the ease that the X y so we have to x e to the three x y and then times three x y plus one. So then this is going to be our partial with respect to why, okay?

So if we want to take the partial this perspective of X and y, remember what we're going to do? So if we were to start off by doing the partial of this respect to x o Dell by Dell X, What we assume is all of these wise here are going to be constant. So just like how we would see is like five or three or eight as a constant. Every why we see here we're going to treat just like it was a constant. And we're just taking the derivative of this with respect to X like we did in single variable calculus. So now we can come over here and so on the left. Since we're doing the partial with respect to X weaken, right, this is F partial X, which we normally just right is the subscript X and then over here. So we just take the derivative just like we would normally for X. And remember, we just treat why, as some constant so three x to the fourth abuse powerful. So it be 12 x cute. Now, since why is it constantly Cuba constant and then multiplied by a number that's still a constant. So the derivative of to like with respect to X for the partial that is going to be zero. Now, over here. Just like how we would pull out this negative seven. We can pull off the why to the fifth and then just take through of Why Square? So the seven. Why? To the fifth. And then we take the derivative of X squared, which is going to give us two X and then plus so eight x squared. Just use powerful. So 16 x three wise. So a constant. Why times another constant is still going to be a constant zero. And then negative. I was just a constant. So plus zero, Then we could go ahead and clean this up a little bit. So are partial with respect to X is going to be 12 x cubed. And I'm just going to move the X out front of the why on then multiply. The two and seven will be negative 14 x y to the fifth plus 16 x. So this here is going to be our partial with respect to X Now, to get the partial with respect to why we're going to do the same thing. So let me just go ahead and stand a little bit. So we're going to do Del by Dell. Why? And now all of these X is we're going to treat as if those air constants. So we would right f partial with respect to why and this is going to be equal to. So if we have a constant raised to a power multiplied by a power will still be a constant. So this will be zero and then two week we just use powerful, so minus six y squared. Um, so again, seven times expert, that's a constant. So we pull that out and then just take the derivative of why to the fifth, which would be five y to the fourth. Then we take the derivative of so constant square times a constant so constant, so plus zero negative three y that just the swift negative three and then the derivative of negative I will just zero that we could go ahead to clean this up so or partial with respect to why is going to be so negative. Six y squared negative 35 x squared y to the fourth minus three. And so then this here is our partial with respect to why


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