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

$$27$$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 10

Related Rates

Derivatives

Missouri State University

Campbell University

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 have been given equation y equals X cubed minus three x and we want to find the value of D Y d t went excess to and DX DT equals three. So a few observations before we actually solve this problem first, when we take these derivatives, you can see in both cases that we're doing this with with respect to t. Now, tea is not one of our equations are one of our variables here. So when I take the derivative of why with respect to T or X with respect to t things like implicit differentiation if I take the derivative, let's say I've got an X cubed here. When I take that derivative that's going to be three x squared on. I have to multiply by dx DT because by saying that I want this with respect to t, I have to use the change rule to add on a dx DT when I'm doing why I'm gonna have to add on d Y d t. Okay, now this is not actually part of the problem, so I'm just gonna erase that. Another question that gets asked often is when do we substitute like, you know, I know I'm looking at when X equals to Do I substitute that in right now? The answer is no. We're not going to do any of the substitution until we find the derivative. You want to find the derivative off the function as a whole just the way it is after we have taken the derivative. That's when we'll plug things in and substitute. So let's take the derivative of our function with respect to t. So the derivative of why is gonna be one d y d t x cubed? That's three x squared times dx, DT, and then minus three times one d x d t. Okay, now I have substance. I've done the derivative. Now is the time to substitute D. Y d t is what I'm trying to find. So that's in the stay. Put everything else that can substitute for Well, X is two. So X squared is going to be four dx DT is three and I'll plug that in in both places. So this is going to equal off three times. Four times three. That's 36 minus nine, which gives me a value of 27. So we found the derivative with respect to T and D Y. D. T. In this case is 27

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