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Find $d y / d x$ using any method.$$5 x^{2}+6 x^{2} y^{2}=y^{2}+15$$

$$\frac{x\left(5+6 y^{2}\right)}{y\left(1-6 x^{2}\right)}$$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 8

Implicit Differentiation

Derivatives

Campbell University

Oregon State University

Harvey Mudd College

Idaho State University

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've been given a function five X squared plus six x squared y squared equals y squared plus 15 and we need to find the derivative of of our function. The easiest way to do this is implicit differentiation. Since I have X's and y's mixed up all over the place here. The thing to remember with implicit differentiation is that why is a function of X? So when I go to take the derivative, let's say I'm taking the derivative of y squared well derivative that I bring down the exponents. Subtract one. It's too. Why? But why isn't just why? Why is a function of X? It stands for something. So the chain rule says I would then have to multiply by the derivative of that piece. So every time we hit a why here when we're taking the derivative, we're gonna put that d Y d x on to whatever else the derivative is to show that we recognize that why is a function of X Okay, So let's take our derivative if well, this first term just has an ex, we're gonna do this just the way we've done everything up to this 0.10 x. Okay, now we have a products. I'm gonna have that six out, and I'm just gonna take the use the product rule on X squared y squared. That's the first times the derivative of the second, which is to y d y dx plus the second times the derivative of the first. And then I have y squared. So again, that's two. Why D y d x, and that's gonna be a plus zero. Okay, so let's get rid of our parentheses. 10 x 12 X squared Y d y d x plus 12 x y squared equals two Y d Y d x. Okay, now, from this point on, this is just algebra. That's all that we're gonna be doing here we're gonna be solving for d Y d X. So I'm gonna pull everything with D I d xto one side, everything without to the other. Since the left side already has all of the non d y DX is, I think I'm gonna leave them there. Everything with a D Y DX is going to go to the other side. Okay, Now, every single term has an even coefficient. So I am just going to divide each of these by two just to simplify things make my life a little bit simpler. Okay, I have put d Y d x on one side. So now I'm gonna factor it out. That gives me a five x plus six x y squared. And if I factor out a d y DX, I'm left with one. I'm sorry I'm left with. Why my apologies There. Why minus six X squared. Why? So I divide both sides by what's in those parentheses and that gives me d y DX equals what's on the left hand side, divided by what's in parentheses. Okay, there's your answer. Now we can do a little bit of simplification here. We can factor out, um, common terms, top and bottom. So top both those terms, haven't X and both terms on the bottom Have a why, and it's not a whole lot simpler, but it's kind of a nice thing to do to factor out when you can. So there's my derivative de y dx off our original function

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