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Math Help - Deriving a square root under a square

  1. #1
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    Deriving a square root under a square

    Suppose that f(x) = \sqrt{6+\sqrt{8x}}. Find f'(x).

    oh man I really need help on this one.
    Last edited by pseizure2000; September 28th 2007 at 05:28 PM.
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  2. #2
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    Maybe if the numbers/expressions/surds are shown, your questions can be solved.

    Two red x's don't help much.
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  3. #3
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    fixed. not sure why i could see it
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  4. #4
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    Quote Originally Posted by pseizure2000 View Post
    Suppose that f(x) = \sqrt{6+\sqrt{8x}}. Find f'(x).

    oh man I really need help on this one.
    Oh there they are!

    f(x) = sqrt[6 +sqrt(8x)]
    Or,
    f(x) = [6 +(8x)^(1/2)]^(1/2)

    f'(x) = (1/2)[6 +(8x)^(1/2)]^(-1/2) *[0 +(1/2)[8x]^(-1/2) *8]}
    Then just simplify that.

    f'(x) = 1 /(2sqrt[6 +sqrt(8x)]) *[4 /sqrt(8x)]

    f'(x) = 2 /(sqrt[6 +sqrt(8x)]*[sqrt(8x)])

    --------------------------------------------------------------------------

    f'(x) = 2 /(2sqrt[1.5 +sqrt(2x)]*2sqrt(2x)) <------ EDIT: This is wrong.

    f'(x) = 1 /[2(sqrt[1.5 +sqrt(2x)] *sqrt(2x)]

    f'(x) = 1 /[2sqrt[3x +2x*sqrt(2x)]

    f'(x) = 1 /[2sqrt[3x +sqrt(8x^3)] ---------------answer
    .

    -----------------------------------------------------------------------------
    Revision:

    f'(x) = 2 /(2sqrt[1.5 +sqrt(x/2)]*2sqrt(2x))

    f'(x) = 1 /[2(sqrt[1.5 +sqrt(x/2)] *sqrt(2x)]

    f'(x) = 1 /[2sqrt[3x +2x*sqrt(x/2)]

    f'(x) = 1 /[2sqrt[3x +sqrt(2x^3)] ---------------revised answer.
    Last edited by ticbol; September 29th 2007 at 03:01 AM. Reason: sqrt(8x) = 4sqrt(x/2)
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  5. #5
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    Quote Originally Posted by pseizure2000 View Post
    Suppose that f(x) = \sqrt{6+\sqrt{8x}}. Find f'(x).
    y=f(x)=\sqrt{6+\sqrt{8x}}\implies y^2=6+\sqrt{8x}, we have

    \begin{aligned}<br />
2yy'&=\frac4{\sqrt{8x}}\\<br />
yy'&=\frac2{\sqrt{8x}}\\<br />
y'&=\frac2{\sqrt{8x}\cdot\sqrt{6+\sqrt{8x}}}<br />
\end{aligned}

    I leave you the make-up.
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