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Finding the nth derivative

by christen1289
Tags: derivative
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christen1289
#1
Sep30-07, 05:58 PM
P: 5
1. The problem statement, all variables and given/known data
Find the formula for the nth derivative of the equation f(x)= 1/(1-x)^2


2. Relevant equations



3. The attempt at a solution
I have no idea how to attempt this problem. I've tried finding derivatives in order to find a pattern but I can't seem to come up with a pattern that would help me to find a formula
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dynamicsolo
#2
Sep30-07, 06:04 PM
HW Helper
P: 1,662
Quote Quote by christen1289 View Post
1. The problem statement, all variables and given/known data
Find the formula for the nth derivative of the equation f(x)= 1/(1-x)^2

3. The attempt at a solution
I have no idea how to attempt this problem. I've tried finding derivatives in order to find a pattern but I can't seem to come up with a pattern that would help me to find a formula
Let's start by listing the derivatives you've found. To make this easier to deal with, you could write the function as

f(x) = (1-x)^(-2) and use the Chain Rule. What is f'(x)?
christen1289
#3
Sep30-07, 06:09 PM
P: 5
f'(x)=2(1-x)^-3

genneth
#4
Sep30-07, 06:14 PM
P: 980
Finding the nth derivative

Can you write that in terms of the original f? Does that help when you apply the derivatives again?
christen1289
#5
Sep30-07, 06:24 PM
P: 5
By finding up to the fourth derivative I came up with this formula:

nth deriv of f= (n+1)(n!)(1-x)^-(n+2)
dynamicsolo
#6
Sep30-07, 06:40 PM
HW Helper
P: 1,662
Quote Quote by christen1289 View Post
By finding up to the fourth derivative I came up with this formula:

nth deriv of f= (n+1)(n!)(1-x)^-(n+2)
Yes! (I had to revise something I was going to say: the (-1) factor from the Chain Rule keeps canceling the minus sign from the exponent-factor, so this does stay positive.)

The one further simplification you can make is that (n+1) (n!) = (n+1)!


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