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Find the differential of each of the given functions.$$y=x(1-x)^{3}$$

Calculus 1 / AB

Chapter 24

Applications of the Derivative

Section 8

Differentials and Linear Approximations

Derivatives

Campbell University

University of Michigan - Ann Arbor

Boston College

Lectures

03:09

In mathematics, precalculu…

31:55

In mathematics, a function…

02:36

Find the differential $d y…

04:21

Solve the given differenti…

01:10

Solve the differential equ…

06:28

00:49

Verify that the following …

02:17

Find $L y$ for the given d…

03:41

Find the general solution …

01:17

08:08

01:35

So in this problem, we're looking at two equations. Why equals one over X to the third minus one and we're looking at why equals one minus X to the third over, two minus X. We're looking for the differentials, right? So all we're really doing here is taking the derivative and rewriting our answer. So in this case, the first thing I want to do with this first problem is rewrite. It's a taking. The derivative will be a little bit easier in this case. I'm gonna rewrite this as X to the third minus one to the negative one. Power taking the derivative of both sides. I've got D Y d X is equal to negative one times X to the third minus one times the derivative the inside, which is three x square. Clean this up. I'm going to get d Y is equal to negative. This should be to the negative. Second power Negative three x squared all over X to the third minus one squared. We got multiply times the outside, subtract one of the expo and take the derivative the inside D X. And so here's my differential for the first problem for the second problem. It's a little bit harder. Teoh, clean that guy up. So I'm gonna have to use the quotient room. So do I. D X equals low de high minus high de lo over. Lolo cleaning that up a little bit, do you? Why, TX, I've got negative six X squared plus three x to the third negative times. A negative is a positive plus one minus X to the third all over two minus X squared, which will give us a final answer of D Y is equal to Chu X to the third because these two are like terms minus six x squared plus one over Tu minus X squared DX here the two differentials for that.

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