Results 1 to 7 of 7

Math Help - Problem with Diff'ing a rational expression

  1. November 18th 2007, 08:17 AM #1
    BClary
    BClary is offline
    Newbie
    Joined
    Nov 2007
    Posts
    10

    Problem with Diff'ing a rational expression

    Keep getting all twisted up in the algebra in finding the first and second derivative of the following

    x(third)-4x(squared)/(x-2)(squared)

    F’(x)
    (x(third)-4x(squared))’(x-2)(squared)-(x(third)-4x(squared))((x-2)(squared))’/(x-2)(fourth)

    than

    (3x(squared)-8x)(x-2)(squared)-(x(third)-4x(squared)(2)(x-2)(1)/(x-2)(fourth)

    than

    (x-2)[(3x(squared)-8x)(x-2)-2x(third)+8x(squared)]/(x-2)(fourth)

    now stuck

    any help on further factoring would be appreciated
    Follow Math Help Forum on Facebook and Google+
  2. November 18th 2007, 08:45 AM #2
    topsquark
    topsquark is offline
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,272
    Thanks
    438
    Awards
    1
    MHF Donor
    Quote Originally Posted by BClary View Post
    Keep getting all twisted up in the algebra in finding the first and second derivative of the following

    x(third)-4x(squared)/(x-2)(squared)

    F’(x)
    (x(third)-4x(squared))’(x-2)(squared)-(x(third)-4x(squared))((x-2)(squared))’/(x-2)(fourth)

    than

    (3x(squared)-8x)(x-2)(squared)-(x(third)-4x(squared)(2)(x-2)(1)/(x-2)(fourth)

    than

    (x-2)[(3x(squared)-8x)(x-2)-2x(third)+8x(squared)]/(x-2)(fourth)

    now stuck

    any help on further factoring would be appreciated
    F(x) = \frac{x^3 - 4x^2}{(x - 2)^2}

    So
    F^{\prime}(x) = \frac{(3x^2 - 8x)(x - 2)^2 - (x^3 - 4x^2)(2(x - 2))}{(x - 2)^4}

    F^{\prime}(x) = \frac{(x - 2)[(3x^2 - 8x)(x - 2) - 2(x^3 - 4x^2)]}{(x - 2)^4}

    This is about where you got to. Now divide through by the common x - 2:
    F^{\prime}(x) = \frac{(3x^2 - 8x)(x - 2) - 2(x^3 - 4x^2)}{(x - 2)^3}

    Expand:
    F^{\prime}(x) = \frac{3x^3 - 8x^2 - 6x^2 + 16x - 2x^3 + 8x^2)}{(x - 2)^3}

    And simplify:
    F^{\prime}(x) = \frac{x^3 - 6x + 16}{(x - 2)^3}

    -Dan
    Follow Math Help Forum on Facebook and Google+
  3. November 18th 2007, 09:05 AM #3
    Soroban
    Soroban is offline
    Super Member
    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,956
    Thanks
    805
    Hello, BClary!

    You seem to be doing fine . . .

    Find the first and second derivatives: . F(x) \;=\;\frac{x^3-4x^2}{(x-2)^2}

    Quotient Rule: . F'(x) \;=\;\frac{(x-2)^2\cdot(3x^2-8x) - (x^3-4x^2)\cdot2(x-2)}{(x-2)^4}

    . . . . . . . . . . . F'(x) \;=\;\frac{x(x-2)^2(3x-8) - 2x^2(x-4)(x-2)} {(x-2)^4}

    Factor/reduce: . F'(x) \;=\;\frac{x(x-2)\cdot\left[(x-2)(3x-8) - 2x(x-4)\right]}{(x-2)^4} \;= . \frac{x\cdot\left[(x-2)(3x-8) - 2x(x-4)\right]}{(x-2)^3}

    Simplify: . . . . F'(x) \;=\;\frac{x\left[3x^2 - 8x - 6x + 16 - 2x^2 + 8x\right]}{(x-2)^3}

    Therefore: . . . \boxed{F'(x) \;=\;\frac{x\left(x^2-6x+16\right)}{(x-2)^3}}


    Now try the second derivative . . .
    Follow Math Help Forum on Facebook and Google+
  4. November 18th 2007, 09:09 AM #4
    BClary
    BClary is offline
    Newbie
    Joined
    Nov 2007
    Posts
    10

    Thanks to all, off doing second now

    Thanks for all your help, lets see where I end up at

    thanks
    Follow Math Help Forum on Facebook and Google+
  5. November 18th 2007, 09:57 AM #5
    BClary
    BClary is offline
    Newbie
    Joined
    Nov 2007
    Posts
    10

    2nd derivative should be

    after a little algrebra (haha)

    (-8x-32)/((x-2)(fourth))

    Right?
    Follow Math Help Forum on Facebook and Google+
  6. November 18th 2007, 09:59 AM #6
    topsquark
    topsquark is offline
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,272
    Thanks
    438
    Awards
    1
    MHF Donor
    Quote Originally Posted by BClary View Post
    after a little algrebra (haha)

    (-8x-32)/((x-2)(fourth))

    Right?
    Yup!

    -Dan
    Follow Math Help Forum on Facebook and Google+
  7. November 18th 2007, 10:00 AM #7
    BClary
    BClary is offline
    Newbie
    Joined
    Nov 2007
    Posts
    10

    Thanks so much

    TY Ty Ty Ty ty TY
    Follow Math Help Forum on Facebook and Google+
« analysis question | l'Hôpital's Rule »

Similar Math Help Forum Discussions

  1. Rational Expression Help
    Posted in the Algebra Forum
    Replies: 6
    Last Post: April 3rd 2011, 06:54 PM
  2. Is this problem considered multiplying a rational expression?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 11th 2010, 03:13 PM
  3. Rational equation, rational expression, simplifying,
    Posted in the Algebra Forum
    Replies: 2
    Last Post: December 8th 2009, 07:48 AM
  4. rational expression
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 9th 2008, 09:31 AM
  5. Rational Expression
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: October 23rd 2008, 01:15 PM

Search Tags


/mathhelpforum @mathhelpforum