Deriving a square root under a square

**pseizure2000** · September 28th 2007, 04:50 PM

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

oh man I really need help on this one.

**ticbol** · September 28th 2007, 05:22 PM

Maybe if the numbers/expressions/surds are shown, your questions can be solved.

Two red x's don't help much.

**pseizure2000** · September 28th 2007, 05:29 PM

fixed. not sure why i could see it

**ticbol** · September 28th 2007, 06:19 PM

Originally Posted by pseizure2000

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.

**Krizalid** · September 28th 2007, 06:31 PM

Originally Posted by pseizure2000

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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