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