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September 18, 2014

September 18, 2014

Posted by **Lost One** on Tuesday, February 26, 2013 at 12:10pm.

- Calc -
**Steve**, Tuesday, February 26, 2013 at 2:44pmy =(x+3)^3 / √(x^2 -1)

lny = 3ln(x+3) - 1/2 ln(x^2-1)

1/y y' = 3/(x+3) - x/(x^2-1)

1/y y' = (3(x^2-1) - x(x+3))/((x+3)(x^2-1))

1/y y' = (2x^2-3x-3)/((x+3)(x^2-1))

y' = (x+3)^3 / √(x^2 -1) * ((2x^2-3x-3)/((x+3)(x^2-1)))

= ((x+3)^2 * (2x^2-3x-3))/(x^2-1)^(3/2)

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