Results 1 to 5 of 5

Math Help - Derive the expression

  1. #1
    Newbie
    Joined
    Apr 2013
    From
    Cork
    Posts
    11

    Derive the expression

    g(x)=x(x^2-1)^1/3
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jul 2012
    From
    INDIA
    Posts
    822
    Thanks
    209

    Re: Derive the expression

    What do you mean by derive? In case you want to differentiate then do it by product rule: ( uv)' = u'v+v'u
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2013
    From
    Cork
    Posts
    11

    Re: Derive the expression

    yes i used product rule but i did not get the answer of 1/3[(5x^2-3)/(x^2-1)^2/3)] which i should have so i am looking for the full workings of this problem
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie MathKnot's Avatar
    Joined
    Apr 2013
    From
    Quebec, Canada
    Posts
    12

    Re: Derive the expression

    Remember, if you look at the function you can see that you need to use both the product and chain rules, so if I use the product rule first, it would be:

    \frac{d}{dx} x((x^2) - 1)^\frac{1}{3} = x\frac{d}{dx} ((x^2) - 1)^\frac{1}{3} + ((x^2) - 1)^\frac{1}{3}\frac{d}{dx} x

    Now you just need to use the chain rule and simplify the resulting expression.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121

    Re: Derive the expression

    Quote Originally Posted by Lozeee View Post
    yes i used product rule but i did not get the answer of 1/3[(5x^2-3)/(x^2-1)^2/3)] which i should have so i am looking for the full workings of this problem
    Then show us what you did! What did you get?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derive the expression
    Posted in the Algebra Forum
    Replies: 5
    Last Post: April 25th 2013, 03:39 AM
  2. Replies: 1
    Last Post: November 29th 2012, 03:59 PM
  3. Replies: 2
    Last Post: March 27th 2011, 08:19 AM
  4. Derive expression based on Cantor SFC
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: September 22nd 2010, 02:14 AM
  5. Replies: 0
    Last Post: November 15th 2008, 06:10 AM

Search Tags


/mathhelpforum @mathhelpforum