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March 31, 2015

March 31, 2015

Posted by **Jacob** on Monday, February 11, 2013 at 9:50pm.

f(x)=(x^2+9)/square roote of x

- Applied Calculus -
**bobpursley**, Monday, February 11, 2013 at 10:02pmf(x)=u/v

f'=u'/v -v'u/v^2

can you do that? u=x^2+9 v=x^.5

u'=2x v'=1/2 x^-.5

- Applied Calculus -
**Jacob**, Monday, February 11, 2013 at 10:09pmYeah, what I accidently did was I tried to solve it further, but I completely over thought that. Thank you Bob!

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