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If $y=\sqrt{3 x^{2}-3},$ find $d y / d t$ when $x=2$ and $d x / d t=-2$.

$$-4$$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 10

Related Rates

Derivatives

Missouri State University

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University of Michigan - Ann Arbor

University of Nottingham

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 an equation. Why equals the square root of three X squared minus three? And what we want to do in this problem is find the value of D Y d t at a given moment. We want X equal to and dx DT t equal. Negative too. Okay, two or three things to point out before we actually start solving this. First you'll notice that we're taking the derivative with respect to t in both cases, which means that both y and X we're treating as a function off t. So when I take the derivative of something for why I have to tack on a d y d t If I'm taking the derivative of something with an ex, I'm going to have to multiply by d x d t to show that these are functions of T. This is a application of the chain rule. Okay, The second thing I want to point out, get rid of those circles. Their second thing is when do we substitute? It's very tempting to say Oh, I know what I what I want extra. I want X equal to let's substitute that in first. No we have to take the derivative first of our function without any substitution. Once we found the derivative, then we can substitute. Okay, so the order is important Derivative first. Then we substitute the last thing I don't wanna point out. And this is just kind of reminder how this works When I have the the square root of something like the square root of X we can think of that is X to the one half. So when I go to take the derivative, that one half comes down, which gives me a two in the denominator, I subtract one from the exponents extra the negative one half, which means I'm putting the square root in the denominator. So any time I have a square root when I'm taking the derivative is going to result in a two times that square root in the denominator. And then I could multiply by the derivative of whatever's under that radical side. So I just wanted Toe says we have a square root here. Just wanted to make sure we remember how to do that. Okay, let's take the derivative of our function. We'll start on the left d Y d t. I have my square root. So my denominator, I have two times that square root in my numerator. I'll take the derivative of what's under the radical that becomes six x times dx DT derivative of negative three is just zero. So I can leave that off. Okay, we have taken the derivative. Now is the time to substitute. I'm trying to find d Y d t so we can leave that everything else we should be able to substitute for I have six. Okay, Excess too. DX DT is negative. Two got two on the bottom here. And if excess to that means this is going to be 12 minus three or nine square root of nine is three. So I've got negative. 24/6 and simplifying. It gives me a value for this derivative of negative four.

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