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September 2, 2014

September 2, 2014

Posted by **Anonymous** on Friday, November 16, 2012 at 11:46pm.

dy/dx=?

- Calculus -
**Steve**, Saturday, November 17, 2012 at 5:51amsince (√(9x^2-1))^2 = 9x^2-1,

I assume you mean arctan^2 √(9x^2-1)

let u = 9x^2-1

y = arctan^2 (√u)

y' = 2 arctan(√u) * 1/(u+1) * 1/(2√u) u'

= 2 arctan(√9x^2-1) * 1/(9x^2) * 1/(2√(9x^2-1)) * 18x

= 2 arctan(√9x^2-1) / [x √(9x^2-1)]

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