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Find $d y / d x$ at the indicated point in two different ways: (a) Solve for $y$ as a function of $x$ and differentiate; (b) Differentiate implicitly.$$3 x^{2}-4 x y-4=0; (2,1)$$

$$1$$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 8

Implicit Differentiation

Derivatives

Campbell University

Harvey Mudd College

Baylor 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 three X squared minus four X Y minus four equals zero, and we're going to take the derivative and evaluated at the point to one. Now, we're actually going to do this two different ways, implicitly and explicitly Implicit is what we have written here, right? My ex is and why It's kind of all jumbled together. Explicit is what we have done up to this point where we have y equals something. And then we take the Y DX and we take the derivative. So let's do that first, since that matches what we've done up to this point. So I'm gonna take our equation that we have three X squared, minus four X y minus four equals zero, and I'm gonna solve for why so everything that is, um I'm gonna move the white term to the right hand side. Everything else is going to stay here. So three X squared minus four equals four x y. And now I'm going to solve for y. It's gonna divide everything by four X, and I kind of like the why on the left hand side. So I'm just going to swap the sides. That gives me three X squared minus 4/4 X. So this is I've explicitly solved this for why, so now I could do t y dx. This is a quotient. So I can use the quotient rule denominator times the derivative of the numerator, which in this case is six x minus the numerator times the derivative of the denominator all over the denominator squared. So let's just simplify this a little bit. That gives me 24 x squared minus 12 X squared plus 16 over 16 x squared and Aiken simplify that. Just a little more 12 X squared plus 16/16 x squared e could make this even smaller, but I think I'm just gonna leave it here for right now because I'm going to plug in our point, going evaluate this at the point to one. So if excess two that's 12 times four, that's 48 plus 16/16 times four is 64. Well, 64/64 gives me a value of one. Okay, Now let's try to do the same thing with implicit differentiation for finding the same derivative from the same function. Evaluated the same space. So our answer better be one. But let's see how that works for implicit differentiation. If I take the derivative of anything with an X, it's just the derivative. So three X square. That derivative will be six x when we get there. However, anytime I take the derivative of why I need to multiply it by D. Y d. X because that's the chain role I'm taking the derivative of why. But that's a function of X, even if I don't know exactly what that function is at this point. So let's begin and take our derivatives derivative of three X squared ISS six x Okay, four x y that is a product. So we have to use the product rule. So I'm gonna just factor out that minus four and I'm gonna take the derivative of X y first times the derivative of the second, the derivative of why is one times d y d x okay. Plus the second times, the derivative of the first well derivative of X is just one derivative of four is zero, and that equals zero. So let's get rid of these parentheses, okay? And now our goal is to solve for D Y d X So everything That's not that I'm gonna push to the other side of the equal sign So I get negative four d y d x and that equals four y minus six x and I'm gonna divide by negative four X and I can simplify this just a little bit. I'm actually gonna break this up. I think, into two pieces. It looks it's fairly simple to do that. So that's gonna give me negative y over X plus three haps. Okay, so let's evaluate at our point, we have the point to one, which means I'll have negative to over one. I'm sorry. Sorry helps if I put the X and the Y in the right spot, so I'll have negative 1/2. Pull us three halves. That gives me two halves or one. So no matter which way we find this derivative, our values are exactly the same.

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