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Find $ dy/dx $ by implicit differentiation.

$ e^y \sin x = x + xy $

$y^{\prime}=\frac{e^{y} \cos x-1-y}{x-e^{y} \sin x}$

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Oregon State University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

in this problem, we are learning how to use the technique of implicit differentiation. Now, when I learned this, I thought that this was very tricky to understand. So hopefully this problem helps you. And this video helps you understand a little bit more about how we use implicit differentiation. So where we start with this equation, eat the Y times sine X equals X plus X y. Now, just looking at this that looks like a really tricky, um equation to find the derivative of How would I do that if I have excess and wise? So the first thing that we're going to do by implicit differentiation is will take the derivative of each side of our equal sign. So we're basically saying that the derivative of E y sine X is equivalent to the derivative X plus X Y. So we'll take the derivative of each side. And when we do that, we'll get E to the Y times d y d x times Synnex plus e to the Y times a co sign of X equal to one plus one times y plus x times D Y d X. Now this is something that we can work with. We have to differentials d Y d X, and what we need is D Y dx by itself. So now every step we make is going to be is going to be in the goal to get d y dx by itself because that is our derivative, so we can do a little bit of simplification. Remember, we're trying to get do y dx by itself so we can subtract this year the y cosine X term. So once we do that, we'll get it to the Y times sine x times do I d x equal toe one plus Why minus e to the y close in x plus x times Do I d. X Now we can do a little bit of factoring to make this easier. We'll have eat the y minus sine x minus X times d Y d x equal toe one plus Why minus eat the y co sign X And now this is a very good position that we're in because this is an entire term times D Y d X, and we want the Dubai DX by itself. So we're just going to divide this term onto the other side to get you I d. X by itself. So we'll get do I d X equals one plus y minus e to the y co sign ex all over Eat the y sine X minus X and that is our derivative. So I hope that this problem helped you understand a little bit more about implicit differentiation, why we use it and how we can go through the process of using it. I know that it's a little bit tricky, but hopefully this made sense and you learn something from it.

University of Denver