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Find $d y / d x$ using any method.$$2 / x-3 / y=x^{2} y^{2}$$

$$\frac{2 y^{2}\left(1+x^{3} y^{2}\right)}{x^{2}\left(3-2 x^{2} y^{3}\right)}$$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 8

Implicit Differentiation

Derivatives

Missouri State University

Campbell University

Baylor University

University of Michigan - Ann Arbor

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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for this problem. We want to take the derivative of two over X minus three over y equals X squared times y squared. Now we have two options. We could solve this for why? And take, uh, the derivative that way, Or we can use implicit differentiation. That's probably our best bet here solving fist for why is not gonna be very easy eyes just much better to leave it all jumbled up. And we could just take the derivative just the way it iss. The key to doing that, though, is to remember that why is a function of X Even though we haven't explicitly said what? Why is it is a function of X. So when I go to take the derivative of something with why, for example, the derivative of y squared well, I'll take the derivative of that bring down my exponents two times y to the first power. But why isn't just a variable? It's a function of X. So the chain rule says we then have to multiply by d Y d X. So every time we're taking the derivative of something with why we're gonna attack that d y dx onto it because of the chain rule to make sure that we have the full correct derivative. So let's go back and look at the equation we were given. I'm going to rewrite this instead of having denominators. I'm gonna bring them up is negative exponents. You can leave them in the denominators and use the quotient rule. I just find this is a little bit easier for me. So that's the way I like to do it. Now let's take our derivative, bring down the exponents and subtract one same thing here. Bring down that exponents, Subtract one. But this is a why. So here's that chain rule D y DX. And on the right hand side we have a quotient. So I'm sorry. Product. So the product rule says it's the first times the derivative of the second, Which is to Why times, D Y. D X plus the second times, the derivative of the first, which is just two X. Okay, so that was the calculus part of this problem. Everything from this point forward is algebra. We're gonna put all the d y DX terms on one side thea other terms to the other and then solve for D Y d X So let's keep the d y dx terms on the left hand side that why? To the negative to I'm just going to drop down to my denominator. Okay, I'm gonna bring the d y DX term over from the right hand side. So that's minus two X squared Y d y d x, And I'm gonna leave the non d y DX terms on the right hand side. So I have to x y squared and I will bring over our other term which will become too over x squared. Okay, factor out D Y d x, and over on the left, I'm left with three over y squared minus two x squared. Why? And on this side, I have two x y squared, plus two over x squared. Okay, Each side has a fraction either. Editors have attracted to a non fraction, so I'm going to do a common denominator on both sides and make it a single fraction on either side. That just makes my algebra a little bit easier in the next step. So on the left, why squared is my common denominator, and I'm gonna end up with three minus two x squared. Why? Cute On the right, Um X squared is my common denominator. And I have two x cubed y squared plus two. Okay, now solving for D Y d X. I have what's on the right to And I actually could factor a two out of here, and I think I might do that. X cubed y squared, plus one over x squared. And I'm going to divide that by what's in the parentheses. Three. Minus two x squared y cubed over y squared. Final step I've got a fraction divided by a fraction. That means I multiply by the reciprocal of that denominator fraction. So that becomes two y squared times x cubed y squared, plus one over X squared times three minus two X squared. Why cute? It's a little messy, but that is the derivative off are given function.

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