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September 23, 2014

September 23, 2014

Posted by **Jesse Stone** on Friday, February 3, 2012 at 9:59am.

- Calculus -
**Steve**, Friday, February 3, 2012 at 11:52amf(x) = x^2 cos^-1(ln(sqrt(x))

f = uv, so f' = u'v + uv'

f' = 2x cos^-1(ln(sqrt(x)) + x^2 (cos^-1(ln(sqrt(x))'

Now, d/dx cos^1(u) = -1/sqrt(1-u^2) u'

u = ln(v), so u' = 1/v v'

v = sqrt(x), so v' = 1/2sqrt(x)

f' = 2x cos^-1(ln(sqrt(x)) + x^2 * -1/sqrt(1-ln^2(sqrt(x)) * 1/sqrt(x) * 1/2sqrt(x)

f' = 2x cos^-1(ln(sqrt(x)) - x/sqrt(4-ln^2(x))

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