Results 1 to 3 of 3

Math Help - Help with derivatives

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    2

    Help with derivatives

    Here is the problem
    Given: Given f(x) = ( sqrt ( x ) - 3) / (sqrt ( x ) + 3 ).

    Find the derivative. I seem to be very weak at my algebra, im not sure what to do after this:

    (sqrt (x) +3 (1/2 x ^-1/2 ) - sqrt (x) - 3 (1/2 x ^-1/2 ) / (sqrt (x) +3)^2 This comes from the quotient rule. Can anyone give me some tips to simplify the equation?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,519
    Thanks
    1404
    Quote Originally Posted by mrge View Post
    Here is the problem
    Given: Given f(x) = ( sqrt ( x ) - 3) / (sqrt ( x ) + 3 ).

    Find the derivative. I seem to be very weak at my algebra, im not sure what to do after this:

    (sqrt (x) +3 (1/2 x ^-1/2 ) - sqrt (x) - 3 (1/2 x ^-1/2 ) / (sqrt (x) +3)^2 This comes from the quotient rule. Can anyone give me some tips to simplify the equation?
    You can either use the quotient rule or simplify the function first.


    Quotient rule:

    f(x) = \frac{\sqrt{x} - 3}{\sqrt{x} + 3}

     = \frac{x^{\frac{1}{2}} - 3}{x^{\frac{1}{2}} + 3}.


    So f'(x) = \frac{(x^{\frac{1}{2}} + 3)(x^{\frac{1}{2}} - 3)' - (x^{\frac{1}{2}} - 3)(x^{\frac{1}{2}} + 3)'}{(x^{\frac{1}{2}} + 3)^2}

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

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

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

     = \frac{3}{\sqrt{x}(\sqrt{x} + 3)^2}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2010
    Posts
    2
    oh i see.. thanks!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivatives and Anti-Derivatives
    Posted in the Calculus Forum
    Replies: 7
    Last Post: February 6th 2011, 06:21 AM
  2. Replies: 1
    Last Post: July 19th 2010, 04:09 PM
  3. Derivatives with both a and y
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 4th 2009, 09:17 AM
  4. Replies: 4
    Last Post: February 10th 2009, 09:54 PM
  5. Trig derivatives/anti-derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 10th 2009, 01:34 PM

Search Tags


/mathhelpforum @mathhelpforum