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Differentiate.

$ f(t) = \frac {sqrt[3]{t}}{1 - 3} $

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00:41

Frank Lin

01:34

Clarissa Noh

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 2

The Product and Quotient Rules

Derivatives

Differentiation

Missouri State University

Baylor University

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.

44:57

In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.

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01:35

All right here we want to take the derivative of this function. F of t cube at t over t minus 3. I'M going to go ahead and rewrite the top as t to the 1 third power and then we're going to go ahead and apply quotient rule. So just a reminder of how cotient roll works, potient rule and basically it helps us take the derivatives of fractions. So let's say i want to in general, have a function that looks like? U f, t over v. F t, then, if i want to find the derivative, then what i'm going to do is take the derivative of the top times the bottom minus the derivative of the bottom times, the top all over the bottom square. All right! So that's potient rule. Let'S go ahead and apply it to our function, so we want to go down here now and go ahead and solve it for our expressions. So primatu is going to be the derivative of the top so that we're gonna use power rule so we'll get 1. Third t to the minus 2 third times the bottom, which is t minus 3, that's a denominator and then i'm going to subtract and take the derivative at the bottom. While the derivative of t minus 3 is just 1 and then we multiply by the top, and then we go all over the bottom square, so that is your derivative. We can take or do our hand at planting it up a little see if it cleans up nice. Sometimes it cleans up nice. Sometimes it doesn't. Let'S tend to take a look and see what we can do to clean this thing up. Okay, so first, let's go ahead and distribute the left term. The 113 of the minus 2 thirds distribute over t minus 3. Let'S you see if that helps us out. If we do that, we'll get 1 third and then t to the 1 third- and let's get my pen to work here there we go okay, so that's the first part of the distribution, then we're going to distribute the minus 3, so the minus 3 times a Third is just 1 and then we'll get t to the minus 2 thirds and then we still have a minus t to the 1. Third, so it looks like we can combine those t to the 1 third, on top all right. Let'S try that, let's see what happens next, we're going to go ahead and just paint up a little bit more again, where we have already solved it, but let's just see half a little fun and clean it up a little bit. I have 1 third to the 1 third minus t to the 1. Third, that will be minus 2. Third, minus 2 thirds t to the 1 third minus t to the minus 2 thirds all over t minus 3 square. So again we have already solved it, but the rest will just be fun kind of tricks to clean it up a bit so 1 way to clean up kind of a messy. A situation with powers, especially negative powers, is to multiply everything top and bottom by um. Basically, the positive version that that negative 1 we're going to multiply everything by t to the 2 thirds and you'll see what happens. If we do this, so it's pretty full and what we'll do is that will help us get rid of any negative um. You know any negative exponents so pretty false: let's go ahead and do that so we're going to distribute this t of the 2 thirds over both parts of the top. So when we do that, we'll get minus 2 thirds and this is cool because i add exponents, so i get t to the 1 third plus t to the 2 thirds and that becomes just t. So that's really pull then, when i multiply to the minus 2 thirds times t to the 2 thirds- and i add the exponents i get to the 0, so i just get minus 1. So that's really cleaned up nicely, but we do have a t of the 2 thirds on the bottom. So i'll go ahead and add it to the bottom and there we have it and the last thing we could do if you want to just plant it up slightly as multiplied top and bottom by 3, and what that does. Is it just gets rid of that fraction in the fraction, and then we have? I guess it looks prettier this way it's no different than the original, but it does have kind of a nice nice compact field to the answer. So this is as cleaned up as i think it's going to get with me and anyway, there was a front problem: how soit helped have a wonderful day?

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