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Problem 78 Hard Difficulty

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

Answer

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

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Video Transcript

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