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Find $d v / d t$ if $v=t \cdot \sqrt[3]{t^{2}-1}$.

$$\frac{5 t^{2}-3}{3\left(t^{2}-1\right)^{2 / 3}}$$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 6

The Chain Rule

Derivatives

Missouri State University

Campbell University

University of Nottingham

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.

02:01

Find $d v / d t$ if $v=t^{…

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00:57

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our task is to find the derivative of this problem, which is gonna require some product oral and the changeable. Because what I would do is rewrite that tu brute as to the one third power to make a problem with a little bit easier. Eso let me show you how I would rewrite that problem. Still have t But then replace this t squared minus one set of the cube root to make it the one third power. So you can identify the product, right? There is the product so you can use the product will and that product rule would say Okay, well, DVD t that notion. Yeah, notation for the derivative, you take the derivative of the left side, which is this one. Eso leave the right side alone, Um, and be able to you if you want to switch back to the cube, root a T square minus one, and then plus, now you leave t alone. And then you take the derivative of the right side, which is bringing that one third in front, uh, t squared minus one. You have to subtract one from your exponents, so that's a negative two thirds power and then times the derivative of the inside, which should be to t uh, now. I wouldn't bother wasting my time simplifying this at all, but some people might rewrite it as the cube root of T squared minus one and then plus C to tee times T would be to t squared over three Q brutes of T squared minus one to the second power eso. This is an acceptable answer as well. Or you could take it all the way down on get the answer of five t squared minus three all over three. And this is just getting the same denominator on both sides t squared minus one to the two thirds power. Any of these answers are acceptable. Uh, it's just where do you want to stop? And I would stop at the beginning. That's just me.

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