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Use the method developed in the previous exercise to find the equation of the line tangent to $x^{2} y^{2}-4 x y+7 y^{3}=5$ at (2,-1).

$$y=-\frac{8}{5} x+\frac{11}{5}$$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 10

Related Rates

Derivatives

Campbell 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 are referencing the results of the previous exercise. Number 27. If you haven't looked there, take a look at 27 1st pause our video here and then come back. So what we saw in Problem 27 was a different way to write the derivative de y DX. If we imagine x and y both as functions of some um unknown. Ah, variable t weaken Say that D y dx is d Y d t over dx DT. Okay, so we're gonna use that result in this problem. We're going to take the derivative and don't have X and y both be functions of tea. So we have a product at our first term. That's the first times the derivative of the second plus the second times, the derivative of the first it. Now we have another product. So again, first times the derivative of the second plus the second times, the derivative of the first. Oops. I'm sorry I wrote the wrong denominator there. That's d t all age of DTs. Okay, our third term. Well, I just bring down that exponents 21 y squared d Y d t and the derivative of a constant is zero. Okay, now what I want to do is create this fraction d y d t over the x DT. So I'm going to combine all of the D Y terms on one side and all the D X terms on the other. So just to help, because there's a lot here. I'm going Thio Use color to group thes three d y d t terms into dx DT terms so d wise I have two x squared y minus four x plus 21 y squared. And there's my d Y d t moving things to the other side. I have positive for why minus two x y squared dx DT. And now I want to make this ratio here. I wanna have d y d t over dx DT because that's going to give me d Y d X. That's the point of doing this. Okay, so, using what I have to make this ratio, I'm going to end up with four y minus two x y squared, divided by two x squared y minus four x plus 21 y squared. Okay, that's a pretty nasty equation, but we're going to use that to find the slope because what our goal is we want to find the tangent line to this curve. So to find the tangent line, we can use point slope form. Why? Minus why one equals m times X minus X one. Our point was given as the point to negative one and our slope we confined by taking our derivative at the point to negative one. So let's substitute in four times. Negative one. That's gonna be negative. Four minus two times x, which is to why squared is one my denominator to while X squared is four. Why is negative one OK, minus eight plus 21. So if I look at this, that gives me negative four minus four. That's a negative eight on top. Negative. Eight minus eight. Um, plus 21. That gives me a positive five. So my slope mm is negative. 8/5. Now we can plug everything in. Why? Minus negative? One plus one equals negative. Eight. Fifth X minus two. And now it's just a matter of simplifying. We get rid of my parentheses and I'm going to subtract one. So when I subtract one common denominator of 5 16 minus five is 11. So the tangent line to our curve at our point is why equals negative 8 50 X plus 11 5th so

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