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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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} $
I still don't understand where the [(Squareroot of 6) / 2 (Square root of X)] comes from.
Originally Posted by Brazuca I still don't understand where the [(Squareroot of 6) / 2 (Square root of X)] comes from. $\displaystyle \sqrt{6x}= \sqrt{6} \sqrt{x}$ so $\displaystyle \frac {d}{dx} \sqrt{6} \sqrt{x} = \sqrt{6} \ \frac {d}{dx} \sqrt{x} = \sqrt{6} \ \frac {1}{2} \frac {1}{\sqrt{x}}$
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