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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.$2 x^{2}-3 x y+4 y^{2}-5 y=2; (1,2)$(Hint: Use the quadratic formula to solve for y.)

$$1 / 4$$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 8

Implicit Differentiation

Derivatives

Missouri State University

University of Michigan - Ann Arbor

Boston College

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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this problem. We've been given an equation that we want to find the derivative we want to find dy dx. We're going to do this two ways. We're going to do this explicitly where we solve this for why and then take the derivative and then we're gonna use implicit differentiation. And we'll find that no matter which way we do this, we will end up with the same value for the, um for the derivative at are given 0.12 Now, let's do our regular explicit differentiation. First, we're gonna solve for y and then, um and then take the derivative like we have up to this point. Now this you're going to see how nice implicit differentiation is because this is not a great problem to solve for why, First, we're going to set this up as the quadratic formula. So I have y squared. Then I have some white pieces, so I'm going to say this is Plus, I have negative three X minus five y, and for a constant, I have two X squared, minus two, equaling zero. So, what I have here, I have a B and see Okay, so we need the quadratic formula to solve this for why So remember quadratic formula says Why equals the opposite of B. So the opposite of R B is going to be three x plus five plus or minus the square root of B squared Negative three x minus five squared. It's B squared minus four times a times C all over to a, which is eight. So let's tidy this up a little bit. Well, underneath this radical if I expand out that squared binomial, that gives me nine X squared plus 30 X plus 25. And if I get rid of these parentheses, it's minus 32 X squared plus 32 all over eight. And let's clean this up even a little bit more. Under that radical, I've got some like terms I can combine. I have negative 23 X squared plus 30 X plus 57 over eight. I'm just gonna put this in a little box here to hold on to it. Now I see that I have a plus or minus here, and when I go, if I take the derivative of this, I'm going to end up with two values. One for the plus one for the minus. I know that my point is +12 If I plug in one for X and two for why the this is actually going to be a plus at the point at which we're looking. We're looking at the plus value because that plus, we'll make this a true statement at the 0.0.12 So I'm going to take the derivative of this equation right here. So I have dy DX. My denominator is just a constant. I'm going to pull out 1/8 okay, and take the derivative of what's left. So I have the derivative of three X, which is three derivative of a constant is gone, and then I have the derivative of a square root. So when I take that derivative that puts a two and the radical in the denominator, remember, you can think of that radical as up to the one half power. So as I bring down that radical, subtract or bring down the exponents, subtract one. You can see where I've gotten this from, and now the numerator is going to be the derivative of what's under the radical, So that's going to be negative. 46 x plus 30. Okay, so that's my derivative. Now, let's evaluate it at the point X equals one, because that's the point that we were given the 10.0.12 Okay, so let's plug in our values. I have three plus okay, on the top. If I let x equal one, that's going to give me a negative 16 on top. In my denominator, if I put X equaling one, I'm going to get the square root of 64 which is 82 times eight is 16. So that's going to be three minus one inside those parentheses or two, 1/8 of two is 1/4. So that's the value of this derivative doing this the normal way. Explicit differentiation solving for y taking the derivative. But what if I do it with implicit differentiation and I think I'm gonna have to come down here, actually, so I'm gonna I'm gonna scroll and I'm gonna re copy my equation. It's two x squared minus three x y plus four y squared minus five. Y equals two. And now I'm going to take the derivative with my exes and wise all jumbled up. The thing to remember with implicit differentiation is why is a function of X? So when I take the derivative of anything with a Y in it, I have to tack on a dy DX on the end. Because of the chain rule, I have to show that Y is a function of X by taking the derivative of why with respect to X. So let's take the derivative of this equation. Well, first term is just with an X that's going to be for X now. I have a product, so I have to use the product rule. So that's the first times the derivative of the second. So that's going to be just dy dx. There's that chain rule. Hey, so the first times the derivative of the second plus the second times the derivative of the first okay, next term just has a Y in it, so it's going to be eight y And again, that chain rule says dy DX minus five dy dx. And that's going to equal zero because the derivative of any constant zero Okay, so let's combine, solve and see what we know here. Um, actually, probably the easiest thing to do instead of solving for dy DX at this point. Let's substitute. I am at the 0.12 So if I plug those in this first term is four minus three dy dx. Why is to 70 minus six plus 16 dy dx minus five dy dx. Okay, if I combined together all of my terms with dy DX, that's gonna be 16 minus eight. That's eight dy dx four minus six is negative. To move that to the other side becomes a positive too. N dy dx is 1/4 That matches what we saw before. So no matter which way we take it, the derivative we will end up with the same value of 1/4.

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