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Suppose that $Q=x^{2} y^{-3} .$ If $d Q / d t=6$ and $d x / d t=2,$ find $d y / d t$ when $x=2$ and $y=1$.

$$1 / 6$$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 10

Related Rates

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 Q equals X squared times. Why, to the negative third. Our goal for this problem is to find D Y d T, given some information at a certain instance, were given a certain X and Y value and values for two of our other rates. Now a couple things to point out before we start solving this first. Every one of thes derivatives is with respect to t. So with us telling us is that Q x and Y are all functions off T now in related races, often time, but it doesn't have to be. It's just some other variable t. So when we take these derivatives, we're gonna have to multiply by. For example, I'm taking the derivative of X squared. When I take that derivative, it's gonna be two x times d x d t X is a function of t, So the channel says I need to multiply this derivative that I've taken by the derivative of X with respect to t. I'm acknowledging the fact that X is a function of T and we're going to do that for every variable Q x and y Okay, the other thing that's often people often get confused on is when do we substitute? I have all these lovely values waiting here. When do I put those in? Do I put them in before I take the derivative? No, we don't. We want to take the derivative first. We want this derivative to be the general derivative of the equation, not at a specific point, just in general. Once I have the derivative, then I can substitute to find what the value is at a particular point. But we have to find the derivative first, so let's go ahead and do that. Let's take the derivative of Are given Equation. Okay, well, on the left, I have D. Q. D T. And then I have a products we need the product rule. That's the first times the derivative of the second. So negative three y subtract one becomes the negative fourth times d Y d t. There's a chain really chain role we were just talking about, plus the second times the derivative of the first, which is going to be two x times dx DT again using that chain rule. Okay, now we've taken the derivative weaken substitute. We want d Y d t. So we're going to substitute in values for everything else. Dick DTs six X is too. So that's going to be four. If why is one? Then why do the negative? Fourth is still one that becomes negative. Three D Y d t is what we're looking for, so I don't have a value for that. Why, to the negative third? Well, if why is one that is one excess to this? Next piece is four and DX DT is too. So what I have ISS six equals negative 12 d Y d t plus h And now it's a matter of solving. We subtract eight from both sides and my last step is to divide by negative 12, and that gives me a value of 1/6.

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