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Given f(x,y) = ln (2xy2 _ 4y) Find the first order partial derivative with respect to yZxy- 2 1 32yZxy - 2 2w2_ 4yAxy-2 2wy- - 4yZxy-1 xy -2y...

Question

Given f(x,y) = ln (2xy2 _ 4y) Find the first order partial derivative with respect to yZxy- 2 1 32yZxy - 2 2w2_ 4yAxy-2 2wy- - 4yZxy-1 xy -2y

Given f(x,y) = ln (2xy2 _ 4y) Find the first order partial derivative with respect to y Zxy- 2 1 32y Zxy - 2 2w2_ 4y Axy-2 2wy- - 4y Zxy-1 xy -2y



Answers

Find the first partial derivatives with respect to $x$ and with respect to $y .$ $$ g(x, y)=\ln \left(x^{2}+y^{2}\right) $$

Mother in this problem, we were asked to find e second order partial derivatives of this function here. So let's begin with the first order. Partial derivatives will start with our sub X. So we're treating excess. The variable y is just a constant. We have natural log of this inter function here, so we follow the chain rule. So we start from the outside in the derivative of log, you know, is one over that inter function and then we need to multiply by the derivative of the inter function. But when we say derivative, we mean partial derivative with respect to X. In this case, though, the partial with respect to acts of X plus y is just one. So we end up with one over X plus y and similarly for our sub y. You get the exact same thing for the same reason. Derivative of log is one over. You have the experts. Why, If we multiplied by the partial with respect to why have exposed why let's again just one. So we get the same things. Okay, so now we're ready to take Theseventy order partial derivatives. Let's start with our ex ex. We just look at our so vex and take its derivative with respect to X. So we want the partial derivative with respect to X of one over X plus. Why? And you could use the quotient rule if you want or think of this as explosive wide of the minus one. Either way will get did the minus one comes out in front by the chain rule power becomes a minus two and we multiplied by the partial derivative with respect to acts of X plus y, which we already said was one. So we just end up with minus one over x plus y all squared. So that is our sub x x our sub y y you can see the same thing is going to happen. We want the partial derivative with respect to why of one over explosive. Why, since that was our one, um, again, we're going to have a partial respect. Why rewrite? Follow the chain rule. And again the exact same thing is going to happen because the derivative of X plus y, with respect to why is again one. So we have the same two answers here. Okay, now, just for the mixed partials are sub x y we could do this all again, except you can see what's gonna happen. Our sub x y Remember, our sub X was the same as our sub y. So this is literally going to be the exact same things. What we just did our some Why? Why? Because our Y and R X were the same. So, um, there's really no need to do the work again because we know it'll be the same calculations and for the same reason are several. I ax will also be the same. So in this case, everything was perfectly symmetrical in every possible way. And, uh, we have all four second order partial derivatives.

Here we have a chav X y equals E to the negative X squared plus y squared. Let's calculate the partial derivative of H with respect to X. So here we have e to the negative X squared plus y squared. And then we still need to multiply. Bye the derivative of negative X squared. So we need a multiplied by negative two X and that's it. Yeah, let's go ahead and figure out the derivative of H with respect. Why? So we get E to the negative X squared plus y squared, multiplied by the derivative of negative y squared. So negative two y and that's it.

Here we have F of x y is equal to x y divided by x squared plus y squared. Let's go ahead and calculate the partial derivative of F with respect to X. So the derivative of the numerator is going to be why and now we multiply that by the denominator, subtract the derivative of the denominator times, the numerator divided by the denominator squared, and I we can simplify a little bit to B y cubed minus x squared. Why divided by X squared plus y squared squared? Okay, let's go ahead and calculate the partial derivative of F with respect to why so First we take the derivative of the top, which is X. Multiply it by the denominator, then subtract the derivative the denominator, and then multiply it by the numerator divide by the denominator squared and the simplifies to X cubed minus two X Y squared, divided by X squared plus y squared, all squared and that's it.

Mother in this problem were asked to find all the ah second order partial derivatives of this function. So let's begin with the first order Partial derivatives w sub X we're treating Expect the very blood of why, like a constant. So this why in front of everything is just a constant that will stay there. The derivative of E to this function is e to that same function now by the chain rule, you need to multiply by the derivative of this function Up here in the exponents, by derivative, remember, would mean the partial derivative with respect to X and just looking at this the derivative of X squared minus y with respect to X is just two ex. Since that, why is a constant So we'll end up with two ex Why e to the X squared minus y Okay, now time for W sub y. This time I'm wise treated as a variable. Now X is the constant. So if we look, we have why times e to the X squared minus why we're going to need the product rule which we didn't for W. Cybex. So the product rule tells us derivative with respect to why will be the first function? Which is why times the derivative meaning the partial derivative respect. Why of the second function? I'm halfway there, plus the derivative of the first function times. The second function. That's the product rule where all are derivatives air partials with respect to why. All right, so let's go ahead and compute this The partial with respect to y Edo The X squared minus y it'll be e to the X squared minus y and I we need to multiply by what's the derivative of X squared minus y with respect to why that's negative one. This is the chain rule in effect, as always. All right, next term, the derivative of why, with respect a wise one in this part stays. So at the end of the day, we have minus Y e to the X squared minus y plus e to the X squared minus y. Which miners will rewrite if we, uh since we can let's pull out that eat of the X squared minus, Why might leave us a negative y and a plus one In other words, one minus y, so they can see what we did. These air our first order partial derivatives. Now we'll use them to get thes second order partial derivatives. So let's start with W sub X X. Once again, we're going to need the product rule. So looking at wciv X, we have, uh, two X Y has one function of X times e to the X squared minus y. That's another function of X, so time for a product rule. Again, we have the first function times, the derivative of the second function. So we should be able to do this without writing out all the steps. Now what's the derivative of this red part? We know it's e to the X squared minus Why times the derivative of X squared minus y with respect to X, which we know is two X Okay, so halfway done with the product rule. We did the first times the derivative the second. Now we have to do the derivative of the of the first on the derivative of two x y. With respect to X is to y times the second function as it is. Okay, so let's clean this up if we can. Two times two is four x times x x squared we have? Why e to the X squared minus y plus two. Why you the X squared minus y? And we can simplify that a bit. We can pull out. Let's see if we can pull up to in poor little why and we could pull out that e to the X squared minus y. All that's left of the first term will be a two X squared, and all that's left of the second term is a one. So there's one of our four final answers. Now let's move on to W sub. Why? Why w sub wise down here. Notice. Once again, we're going to need a product rule. We have a first function with a wine at times a second function with a Y in it. So let's use the product rule again. We need the first function times, the derivative of the second function with respect to why now the derivative of one minus y, with respect a Why is negative, plus the derivative of the first function. So what's the derivative of E to the X squared minus y? Well, it's itself you the X squared minus y times the derivative of X squared minus y with respect to why, which is negative. Long Okay, times the second part as it is. So that's the product rule and cleaning things up. We have minus E the X squared minus y minus one minus y times e to the X squared minus y. And we can see we can factor out one of the fact you're minus you to the X squared minus y. And that will leave one plus one minus y. But really, that's two minus. Why? Obviously So will this write it like that to minus y? And there is our final answer for W sub? Why? Why? Okay, now it is time for the mixed partials. So W sub X y looking ads. Uh, this w sub x. It would have to use the product rule again because both these functions have a wine. It Why don't we be a little clever here? We noticed. Look at the function w Every function here is continuous right squaring, uh, exponential Sze Everything else is continuous and all of our derivatives here also continuous so we can invoke Claros theorem fear him to from this section to say that wciv X y equals w sub y X. Because all these derivatives and the original are continuous, the two mics partials must equal each other. This is clever because it's it'll be much easier to take. W sub y X. Look at wciv wide down here now. We don't need the product rule. Only one of these has an accident, so well, let's do it that way. Come in with respect to X now one minus wise, it's a constant so that can remain in there. Were multiplying by constant and the derivative of eat of the X squared minus y with respect to X is eat the X squared minus y times two x. We've seen that before. So final answer. Let's just leave it like this. And these are the last two second order partial derivatives. The two mixed partials, which are equal to each other in this


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