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Use implicit differentiation to find $d y / d x$.$x^{2} y+x y^{2}=6$

$\frac{d y}{d x}=\frac{-2 x y-y^{2}}{x^{2}+2 x y}$

Calculus 1 / AB

Chapter 3

Derivatives

Section 7

Implicit Differentiation

Maneesh B.

October 3, 2019

Just a quick point whenever there is multiplication between two variables you have to use the product rule. Sorry for getting mixed up and saying that it was chain rule.

Missouri State University

Campbell University

Baylor University

University of Nottingham

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s O today will be solving this problem. So this problem is X squared times Why plus X Y squared is equal to six. So when doing the implicit differentiation of this equation don't forget to, uh, do the chain rule for each of these multiplication Sze. So first you want to take the derivative of X squared which will be two x times Why, plus X squared times too dirty of why Which will be one and you don't want to forget uh, including d y over dx cause whenever you take the differentiation of and why variable you want to include ah d y over DX. This is what is known as implicit differentiation So plus the derivative of X which will be one times why square plus x times the dirt of the derivative of y squared Which will be to why times don't forget to put d x again d y over DX. My bad is equal to zero because you have to take the derivative of six which will which is a constant and so become zero. So after that, you want a group all the d y over DX variables together and ah and the rest of the other components in this equation together, So this will turn out to be two x. Why plus y squared plus x squared times D y over d x close to x y comes D Y over The ex is equal to zero And now you want to take all these variables that don't have to be Why, over the X component to the other side So you could single out this d y over dx um, in place so that you get the implicit differentiation of this equation. So it becomes that's a square times D y over DX because to X y times d y over d x is equal to negative to x y minus y squared cause he's a practice this on both sides So it will come out to negative to X uh, why I minus y squared and then you have to cept substitute this do whatever d x variable out of the equation. So we d y over d x times X square crush two x Why is equal to negative to X minus y squared. So if you take the distributive property and you do, um, you distributed to each of these terms, you'll get X squared times do whatever the X plus two x y 10 31 of dxl, which is the original of equation back in this step. So now, um, once using a lot does d y over DX. You take this other other side by dividing on both sides, so turn out to be d y over d X is equal to negative to x y my nice way square over x squared cause to X. Why, when this is a solution for the implicit differentiation of dysfunction. I hope this helped, um, if you have any questions, just reach out to me and, uh, thank you.

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