Suppose that . Find

oh man I really need help on this one.

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- Sep 28th 2007, 04:50 PMpseizure2000Deriving a square root under a square
Suppose that . Find

oh man I really need help on this one. - Sep 28th 2007, 05:22 PMticbol
Maybe if the numbers/expressions/surds are shown, your questions can be solved.

Two red x's don't help much. - Sep 28th 2007, 05:29 PMpseizure2000
fixed. not sure why i could see it

- Sep 28th 2007, 06:19 PMticbol
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)])

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*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

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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. - Sep 28th 2007, 06:31 PMKrizalid