Results 1 to 5 of 5

Math Help - [SOLVED] Deriving a square root under a square

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    31

    [SOLVED] Deriving a square root under a square

    Suppose that . Find

    (moved from urgent and fixed) oh man I really need help on this one.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    South Africa
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by pseizure2000 View Post
    Suppose that . Find

    (moved from urgent and fixed) oh man I really need help on this one.
    I know the thread says solved, but i'll do it anyway so you can check.

    f(x) = (6 + (8x)^{ \frac{1}{2} } )^{ \frac{1}{2} }
    f(x) = 6^{ \frac{1}{2} } + (8x)^{ \frac{1}{4} }
    f(x) = 6^{ \frac{1}{2} } + (8)^{ \frac{1}{4} } \times (x)^{ \frac{1}{4} }

    f '(x) = \frac{ \frac{1}{4} \sqrt[4]{8} }{ \sqrt[4]{x^3} }
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by janvdl View Post
    I know the thread says solved, but i'll do it anyway so you can check.

    f(x) = (6 + (8x)^{ \frac{1}{2} } )^{ \frac{1}{2} }
    f(x) = 6^{ \frac{1}{2} } + (8x)^{ \frac{1}{4} }
    f(x) = 6^{ \frac{1}{2} } + (8)^{ \frac{1}{4} } \times (x)^{ \frac{1}{4} }

    f '(x) = \frac{ \frac{1}{4} \sqrt[4]{8} }{ \sqrt[4]{x^3} }
    But (a +b)^(1/2) is not equal to a^(1/2) +b^(1/2)

    Or more specifically,
    (6 +(8x)^(1/2))^(1/2)
    is not
    6^(1/2) +(8x)^(1/4)

    Example,
    (4 +sqrt(81))^(1/2)
    is not
    4^(1/2) +(81)^(1/4) = 2 +3 = 5

    (4 +sqrt(81))^(1/2)
    = (4 +9)^(1/2)
    = sqrt(13) ----------------------not 5.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,662
    Thanks
    298
    Awards
    1
    Quote Originally Posted by pseizure2000 View Post
    Suppose that . Find

    (moved from urgent and fixed) oh man I really need help on this one.
    f(x) = \sqrt{6 + \sqrt{8x}}

    Write this as:
    f(x) = (6 + (8x)^{1/2} )^{1/2}

    Now use the chain rule:
    f^{\prime}(x) = \frac{1}{2} \cdot (6 + (8x)^{1/2})^{-1/2} \cdot \left ( \frac{1}{2} \cdot (8x)^{-1/2} \cdot 8 \right )

    f^{\prime}(x) = \frac{2}{\sqrt{6 + \sqrt{8x}}\sqrt{8x}}

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    11
    Mmm... see here.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Square root inside square root equation
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 10th 2011, 04:17 PM
  2. Replies: 11
    Last Post: October 25th 2009, 06:45 PM
  3. Replies: 12
    Last Post: November 22nd 2008, 12:41 PM
  4. [SOLVED] square root
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 22nd 2008, 03:09 PM
  5. Deriving a square root under a square
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 28th 2007, 06:31 PM

Search Tags


/mathhelpforum @mathhelpforum