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Find $ dy/dx $ by implicit differentiation. $ \frac {x^2}{x + y} = y^2 + 1 $

$y^{\prime}=\frac{x(x+2 y)}{2 y(x+y)^{2}+x^{2}}$

01:03

Frank L.

01:47

Doruk I.

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 5

Implicit Differentiation

Derivatives

Differentiation

Oregon State University

Baylor University

University of Nottingham

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So for this poem, we're gonna have to use quotient rule to perform inputs of differentiation because what we have is X squared over X plus. Why equals y squared plus one. So I'll take the derivative of both sides when we take the derivative outside's, we have to use the quotient rule on the left side. So it's going to be, uh, the low or the numerator times the derivative of the numerator, so tend to x and then, um, minus X squared times he derivative of X plus y, which we know is going to end up being one plus. Why prying that Obama. I'll be in parenthesis right there. That's what we have going for us. Then, Um, what we'll have is that this is equal. This is all over the denominator squared. So I close, why squared? And then this is equal to We're gonna want Thio do change again. Why Squared will give us two y times y prime plus zero because one is just constant. So this is what we have. Then we can multiply both sides by X plus y squared. So multiply that on both sides and what will end up getting as a result, is this Times X plus y squared. Then once we do that, we can add X squared plus D Y d. X on both sides. And the reason why we can do that is because we end up getting the two x times X plus y minus X squared, minus X squared times. All of this. So when we subtract X squared d y d X from both sides, what will end up getting is a two x squared us two x y minus X squared equal to to why x plus y squared plus X squared times D Y d x or just white prime, then dividing both sides by two y times X plus y squared plus X squared. We will get rid of this right here. And then we will have this all being divided by this value right here. So that is our white crime.

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