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This question's previous answer on Chegg is incorrect.Find the first partial derivatives of the function.f(x, y) = ax + bycx + dy...

Question

This question's previous answer on Chegg is incorrect.Find the first partial derivatives of the function.f(x, y) = ax + bycx + dy

This question's previous answer on Chegg is incorrect.Find the first partial derivatives of the function.f(x, y) = ax + bycx + dy



Answers

Find the first partial derivatives of the function.
$ f(x, y) = \dfrac{ax + by}{cx + dy} $

Were given a function. And we're asked to find all the second partial derivatives of dysfunction. The function is F of X. Y equals the natural log of A X plus B times Y. My adopted japanese father please. Yeah. You see, yeah. Sorry to find the second partial derivatives first, we'll find the first partial derivatives, the partial derivative with respect to X. By the chain rule. I'm going to hold Y constant and differentiate X with respect to ex I mean using the chain rule. So this is going to be a over A X plus B. Y. Likewise, the first partial derivative F. With respect to why I find this by holding X constant and differentiating F with respect to Y. Using the chain rule. And this is the over A X plus B. Y. To find the second partial derivative of F with respect to X. I'm going to hold wide constant in the first part. A very live of F with respect to X. And then differentiate with respect to X. This is negative A squared over a X plus B Y squared. Yeah, you should get work through the house. How much would that cost now to find the partial derivative of the second partial derivative of F. First with respect to X. Then with respect to Y. Well, I'm going to hold X constants in the first partial derivative of F with respect to X. And differentiate with respect to why this gives us negative eight times be over A X plus B Y squared. Hold it up. No, I would have a manager likewise. To find the second partial derivative of F. First with respect to Y. Then with respect to X. I'll hold Why constant in the first partial derivative of F with respect to Y. And then differentiate with respect to X. And I also get negative A. B over a X plus B Y squared. Notice that the two mixed partials are equal. We in fact, know that this is true by Schwartz's theorem or a freedom on the equality of mixed partials because our function F has continuous second partial derivatives at all points in its domain and flying over the stadium now is man and saying. And finally, the second partial derivative of F with respect to why we find by holding X constant and the first partial derivative of F with respect to Y and differentiating with respect to X. Sorry, with respect to Y. And we get negative B squared over A X plus D. Y squared. Well, how

All right. Uh huh. We want to find the first partial derivatives of the function F of X. Y is equal to X. L N Y plus Y L N X. This question is testing our ability to take derivatives of functions or multiple variables with respect to a singular variable, which is important for Multivariate calculus to start off with, let's note what derivatives we want to find. The first partial derivatives were searching for our fx equals DFB X and F Y equals the FBI. And to find these, we note that we simply take the derivative of F with respect to each variable. Where we treat other variables and constants. For fx we treat why are the constant and F Y. We treat X as a constant. So with this in mind we can go ahead and take these derivatives Fx and Fy are given here. So for F X, Y. Or rather fx during X. L n Y is a L N Y. Because Y is constant and rid of Y, L n x is Y times derivative, L n X or Y times one over X, giving L n Y plus Y, or X. Next. For fy, we again have X times one over Y plus L n X equals X over Y plus L n X.

Graphic. We want to find the first partial derivatives of the function F of X, Y is equal to X. Y over X plus Y, which I re written here on the right as E to the X Y times X plus Y to the negative. First. This question is testing our understanding of how to take derivatives of multi variant functions, which is essential for multi variable confidence. The first partial derivatives that were searching for from F r, fx and Fy where fx is just the F B X and F Y is the FBI. We note that when we take each of these derivatives, we treat the other variable and constant. So for F X Y is a constant. And for fy we treat X as a constant. We also know that we need the product rule in order to take this derivative as well as the chain rule. So our relatives are as follows, given this information fx and fy first for fx we take the product rule as I mentioned. So we take the derivative of the first term times second term and add to the first term times during the second term. We see that each of these terms has an exposed wide square denominator. So to normalize this, we multiply the first term here by X plus Y over X plus Y. Which allows us to rewrite our solution as X Y. E. To the X, Y plus Y squared E x y minus the X. Y over X plus Y squared. Next for fy, we see that the derivative is near identical. So again we take the product rule and then we multiply the first term by X plus Y over X plus Y. To get final solution. X, Y e x, Y plus X squared dx y minus E x Y over X plus y square


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