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Find the 1000th derivative of $ f(x) = xe^{-x}. $

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1000 th derivative of $f(x)=(x-1000) e^{-x}$

02:13

Frank Lin

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 4

The Chain Rule

Derivatives

Differentiation

Missouri State University

University of Michigan - Ann Arbor

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.

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.

04:20

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Our goal in this problem is to find the 1/1000 derivative so clearly we're not going to take 1000. Derivatives were going to take a few derivatives and then we're going to find a pattern. So let's start with the first derivative we need to use the product rule. So we have the first X times, the derivative of the second E to the opposite of X times. Negative one. We got that from the chain rule, plus the second each of the opposite of X times, the derivative of the 1st 1 Okay, we can factor e to the negative x out of both of those, and we have each of negative X times one minus X. That's our first derivative. Now let's move on to the second derivative. We're going to need to use the product rule on that so we have the first he to the night negative X times the derivative of the second negative one, plus the 2nd 1 minus x times the derivative of the first E to the negative X times negative one. We can factor negative one times e to the negative x out of both of those terms and That leaves us with one plus one minus X, so that would be two minus X. Now let's find the third derivative. So we're going to use the product rule on what we just got. So have the first theocracy it of eat the opposite of X times, the derivative of the second negative one, plus the 2nd 2 minus X times the derivative of the first, the opposite of E to the opposite of X times. Negative one. Okay, so what we can do here is simplify. Multiply the negatives. We get a positive. So each of the opposite of X plus tu minus X times each of the opposite of X. And if we factor the each of the opposite of Exide. Of those, we have one plus two minus X who we have three minus X. Now we might start to get a glimmer of what the pattern is. Let's do one more. Let's do the fourth derivative. So we have a product rule again. The first heat of the opposite of X times, the derivative of the second negative one, plus the 2nd 3 minus x times the derivative of the first eat of the opposite of X times negative one. Let's factor Negative one times e to the opposite of X out of both terms, and we have one plus three minus X. We have four minus X. Okay, lets see what's happening here. Notice that we have alternating positive, negative, positive negative in front of our each of the negative X. And then we had for the first derivative one minus X for the second derivative two minus X, or the third derivative three minus X. And for the fourth derivative, four minus X. So the 1/1000 derivative would have a negative in front of it, since all the evens seemed to have negatives in front of them and it would have a 1000 minus X. So we end up with the 1/1000 derivative. Beth is the opposite of each, the opposite of X times, 1000 minus X and another way we could write that if we want to distribute the negative through the parentheses is e to the opposite of X Times X minus 1000

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