Question

Find the nth derivative of f(x), where n is an integer.

f(x)=(x^n)/(1-x)

I've attempted it several times, but each time I can't seem to figure out the pattern for the derivative. My guess would be something along the lines of,

f^n(x)=[n!(x^(n-1))]/[(n-1)!(1-x)^(2-n)]

Answer 1

I might try something like this:

\(f(x) = \dfrac{x^{n}}{1-x} = x^{n}(1+x+x^{2}+x^{3} + ... ) = x^{n} + x^{n+1} + x^{n+2}+...\)

What does the nth derivative of that look like?

Answer 2

lol

Answer 3

@theonechewbaca, if you're wondering how @tkhunny got that, it comes from the power series for \(\dfrac{1}{1-x}\), which is \(\displaystyle\sum_{n=0}^\infty x^n=1+x+x^2+\cdots\) (a geometric series, so it only converges for \(|x|<1\)).

Answer 4

@tkhunny, the nth derivative would look like a formula that gives the equation for any derivative taken from the nth derivative equation. Something like the one I have posted originally on the post. @sithsandgiggles, how would i make that into a nth derivative form?

Answer 5

\[
\dfrac{d^n}{dx^n}(x^{n} + x^{n+1} + x^{n+2}+\dots) = \frac{n!}{0!}x^0+\frac{(n+1)!}{1!}x^1+\frac{(n+2)!}{2!}x^2+\dots
\]

Answer 6

This is a very interesting pattern.

Answer 7

Remember that the MacLaurin series is: \[\Large
\sum_{n=0}^\infty f^{(n)}(0)\frac{x^n}{n!}
\]

Answer 8

is that equivalent to x^n/1-x?

Answer 9

Because that is the function I need to figure out the nth derivative of. I'm not really understanding the connection between the two :s

Answer 10

Well it's weird because it's a function where \[
f(0) = n!\quad f'(0) = (n+1)!\quad f''(0)=(n+2)!
\]

Answer 11

I wonder if we're going down the wrong path here.

@tkhunny Got any other hints?

Answer 12

Not at the moment. It was a wonderful exploration.

Answer 13

It becomes a weird diffeq \[
f'(0) = (n+1) f(0)
\]

Answer 14

I say weird because It seems very strange that \(e^{kx}\) would be the result.

Answer 15

So would the nth derivative just be n!0!x0+(n+1)!1!x1+(n+2)!2!x2+…

Answer 16

I'm not completely sure @theonechewbaca

Answer 17

What course is this?

Answer 18

This is from a calc1 course, living in canada

Answer 19

Okay we've been going into math that is a bit over the top for calc 1.

Answer 20

http://en.wikipedia.org/wiki/Leibniz_rule_(generalized_product_rule)

Answer 21

Right, where f(x) could be the x^n, and g(x) could be (1-x)?

Answer 22

Yeah, \(g(x)=(1-x)^{-1}\)

Answer 23

Could you maybe translate the rule into something I could understand? I'm not really sure what to make of the formula on the wiki page.Thanks!

Answer 24

It came to me when I woke up, this morning. Logarithmic differentiation works.

Define \(h(x) = ln(f(x)) = n\cdot ln(x) - ln(1-x)\)

The derivatives of h(x) are trivial and trivially patterned.

Answer 25

Right, so because you introduced the natural log, it splits the equation up and if I differentiate it, would it be easier to see the pattern for the nth derivative equation?