Results 1 to 4 of 4

Math Help - Hard question - nth derivative

  1. #1
    Senior Member polymerase's Avatar
    Joined
    May 2007
    From
    Sydney
    Posts
    267

    Hard question - nth derivative

    f(x)=\dfrac{x^n}{1-x} where n is a constant. Find an expression for the n^{th} derivative, f^{(n)}(x)

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    \displaystyle\frac{x^n}{1-x}=\frac{x^n-1+1}{1-x}=-\frac{1-x^n}{1-x}+\frac{1}{1-x}=
    =-(1+x+x^2+\ldots +x^{n-1})+\frac{1}{1-x}

    Now, (1+x+x^2+\ldots+x^{n-1})^{(n)}=0
    and \left(\frac{1}{1-x}\right)^{(n)}=\frac{(-1)^{n+1}n!}{(1-x)^{n+1}}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member polymerase's Avatar
    Joined
    May 2007
    From
    Sydney
    Posts
    267
    Quote Originally Posted by red_dog View Post
    \displaystyle\frac{x^n}{1-x}=\frac{x^n-1+1}{1-x}=-\frac{1-x^n}{1-x}+\frac{1}{1-x}=
    =-(1+x+x^2+\ldots +x^{n-1})+\frac{1}{1-x}

    Now, (1+x+x^2+\ldots+x^{n-1})^{(n)}=0
    and \left(\frac{1}{1-x}\right)^{(n)}=\frac{(-1)^{n+1}n!}{(1-x)^{n+1}}
    From here:
    Now, (1+x+x^2+\ldots+x^{n-1})^{(n)}=0
    and \left(\frac{1}{1-x}\right)^{(n)}=\frac{(-1)^{n+1}n!}{(1-x)^{n+1}}

    could you briefly explain what you did?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    Let g(x)=1+x+x^2+\ldots+x^{n-1}.
    g is a n-1-degree plynomial and the n-th derivative is 0.

    Let h(x)=\frac{1}{1-x}.
    We have h'(x)=\frac{1}{(1-x)^2}
    h''(x)=-\frac{2}{(1-x)^3}
    h'''(x)=\frac{3!}{(1-x)^4}
    By induction h^{(n)}=\frac{(-1)^{n+1}n!}{(1-x)^{n+1}}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. extremely hard partial derivative question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 20th 2013, 07:29 AM
  2. Hard Derivative.
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 22nd 2009, 09:18 PM
  3. A hard inverse tan partial derivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 20th 2009, 04:12 AM
  4. Replies: 0
    Last Post: February 11th 2008, 03:22 AM
  5. Hard derivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 24th 2007, 03:28 AM

Search Tags


/mathhelpforum @mathhelpforum