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$2 x^{2}+3 y^{2}=14 .$ Determine the equation of the tangent line at each of the following points: (a) (1,2)$;$ (b) (1,-2).

(a) $y=-\frac{1}{3} x+\frac{7}{3}$(b) $y=\frac{1}{3} x-\frac{7}{3}$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 8

Implicit Differentiation

Derivatives

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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 in equation two X squared plus three y squared equals 14, and we want to find the equation of the tangent line at a few different specified points. Now, what do we need to know in order to find the equation of a line? Well, there's a few different ways that we can write a line. We have the slope intercept form, which isn't very helpful here because I don't have the slope or the intercept. But what is helpful is the point slope form. Why minus y one equals m times X minus X one. Because for both of these tangent lines, we wanna find they give us a point. So the only thing I'm missing is the slope. Well, the slope I confined using the derivative. So we're gonna have to find the derivative of this equation and then I can evaluate it. At my point, that will give me both a point and a slope, and I'll be able to write my equation for the tangent line. So let's come back to our equation. We're gonna use implicit differentiation here, uh, with the X squared in the Y squared implicit differentiation just makes a lot of sense here. It's easier, I think, that solving for y and taking the derivative One thing to keep in mind is that why is a function of X So anytime we come across a Why, when we're taking the derivative, we have to tack on that D Y DX because of the chain rule. Okay, why, you know we take the derivative, but we're not done because why isn't just a variable? It's a function. It's even though we don't have it specifically written out. What why equals Why is a function of X? So let's take our derivative the derivative of two x squared. When we bring down that exponents that becomes four x three y squared again, we bring down the exponents that six y times d y DX. There's air chain rules, and that equals zero. And let's solve for D Y d x six y d Y d X, and I'll move that four x to the other side and divide by six. Why, okay, and let's just simplify this a little bit, make the numbers the smallest possible. I could divide the top and bottom, both by two, so there's a derivative. So that's going to give me the slope anywhere around on this equation. So let's look at the points were supposed to evaluate. At first we have the 0.12 Well, what's the slope at the 0.12? Anywhere on this equation? The slope is negative. Two x over three y. Well, if X is one that puts a negative two on top. If why is to that gives me a six on the bottom or negative one third. So now I have Slope. I have a point. Let's plug it into our equation. Why minus two equals my slope. Negative one third X minus one And just to clean it up, let's put this into Slope. Intercept form is the most common way to right our final answer. So I'll get rid of my parentheses and it will add to to both sides well, adding to that's adding six thirds. So I'm gonna end up with negative one third X plus seven thirds. Okay, let's try our second point. That's the 0.0.1 negative too well again, figure out our slope. The equation for our slope is negative. Two x over three Y X is one that gives me a negative, too, if why is negative to that gives me a negative six, so it's almost the same. But my sign has changed, and now my slope is positive. One third. Let's plug those into my equation. Why, minus negative, too, Which is why plus two equals slope times X minus my X coordinate, which is X minus one and just like last time, let's get rid of our parentheses, says one third X minus one third. Subtract two from both sides. That gives me one third x minus seven thirds. So those are two equations for the tangent lines for the given equation at the given points.

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