http://img43.imageshack.us/img43/6099/squarej.jpg

I circled the part I didn't understand when solving this equation.

When I did it I turned U into 6^(1/2) and DU into 3X^(-1/2)

So I understand where the (1/1+6x), but I don't understand the rest.

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- Jul 20th 2009, 06:30 PMBrazucaCalculate Derivative? Square root?
http://img43.imageshack.us/img43/6099/squarej.jpg

I circled the part I didn't understand when solving this equation.

When I did it I turned U into 6^(1/2) and DU into 3X^(-1/2)

So I understand where the (1/1+6x), but I don't understand the rest. - Jul 20th 2009, 06:45 PMRandom Variable
use the first fundamental theorem of calculus and the chain rule

$\displaystyle \frac {d}{dx} \int_{0}^{\sqrt{6x}} \frac {dt}{1+t^{2}} = \frac {1}{1+(\sqrt{6x})^2} \frac {d}{dx} \sqrt{6x} $ - Jul 20th 2009, 07:00 PMBrazuca
I still don't understand where the [(Squareroot of 6) / 2 (Square root of X)] comes from.

- Jul 20th 2009, 07:12 PMRandom Variable