Results 1 to 2 of 2

Math Help - Struggling with quotient rule question

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    13

    Struggling with quotient rule question

    Doing a practise paper question.

    My answer to an earlier part for using composite rule on f(x)=(x^2 -6x+23)^\frac{3}{2 }

    I got \frac{3}{2 }(x^ -6x+23)^\frac{1}{2 } (2x-6)

    Now I need to use quotient rule to differentiate g(x)=\frac{x-2}{(x^2 -6x+23)^\frac{3}{2 }  }

    The working so far is g'(x)=\frac{[(x^2 -6x+23)^\frac{3}{2 }(1)] - [(x-2)(\frac{3}{2 }(x^2 -6x+23)^\frac{1}{2 }(2x-6))   }{[(x^2 -6x+23)^\frac{3}{2 }]^2  }

    g'(x)=\frac{[(x^2 -6x+23)^\frac{3}{2 }] - [(2x^2 - 10x+12) \frac{3}{2 }(x^2 -6x+23)^\frac{1}{2 }]  }{(x^2 -6x+23)^\frac{6}{ 2}  }  -

    g'=\frac{(x^2 -6x+23)^\frac{3}{2 } - [(3x^2 -15x +18)(x^2 -6x+23)^\frac{1}{2 }  }{ (x^2 -6x+23)^\frac{6}{2 } }



    and here I'm stuck with how to reach the answer i have of

    g'(x)=\frac{5+9x-2x^2}{(x^2 -6x+23)^\frac{5}{2 }  }
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member abhishekkgp's Avatar
    Joined
    Jan 2011
    From
    India
    Posts
    495
    Thanks
    1
    Quote Originally Posted by Turlock View Post
    Doing a practise paper question.

    My answer to an earlier part for using composite rule on f(x)=(x^2 -6x+23)^\frac{3}{2 }

    I got \frac{3}{2 }(x^ -6x+23)^\frac{1}{2 } (2x-6)

    Now I need to use quotient rule to differentiate g(x)=\frac{x-2}{(x^2 -6x+23)^\frac{3}{2 }  }

    The working so far is g'(x)=\frac{[(x^2 -6x+23)^\frac{3}{2 }(1)] - [(x-2)(\frac{3}{2 }(x^2 -6x+23)^\frac{1}{2 }(2x-6))   }{[(x^2 -6x+23)^\frac{3}{2 }]^2  }

    g'(x)=\frac{[(x^2 -6x+23)^\frac{3}{2 }] - [(2x^2 - 10x+12) \frac{3}{2 }(x^2 -6x+23)^\frac{1}{2 }]  }{(x^2 -6x+23)^\frac{6}{ 2}  }  -

    g'=\frac{(x^2 -6x+23)^\frac{3}{2 } - [(3x^2 -15x +18)(x^2 -6x+23)^\frac{1}{2 }  }{ (x^2 -6x+23)^\frac{6}{2 } }
    take (x^2-6x+23)^{\frac{1}{2}} out from the numerator and then cancel this with the denominator



    and here I'm stuck with how to reach the answer i have of

    g'(x)=\frac{5+9x-2x^2}{(x^2 -6x+23)^\frac{5}{2 }  }
    ...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quotient rule question.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 12th 2010, 09:20 AM
  2. Quotient Rule - Factorization Question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 28th 2010, 04:31 PM
  3. Replies: 19
    Last Post: October 19th 2009, 06:10 PM
  4. Quotient Rule Question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 26th 2009, 01:39 PM
  5. Derivative question using the Quotient rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 17th 2007, 10:05 AM

Search Tags


/mathhelpforum @mathhelpforum