Calculate Derivative? Square root?

**Brazuca** · July 20th 2009, 06:30 PM

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.

**Random Variable** · July 20th 2009, 06:45 PM

use the first fundamental theorem of calculus and the chain rule

$\frac {d}{dx} \int_{0}^{\sqrt{6x}} \frac {dt}{1+t^{2}} = \frac {1}{1+(\sqrt{6x})^2} \frac {d}{dx} \sqrt{6x}$

**Brazuca** · July 20th 2009, 07:00 PM

I still don't understand where the [(Squareroot of 6) / 2 (Square root of X)] comes from.

**Random Variable** · July 20th 2009, 07:12 PM

Originally Posted by Brazuca

I still don't understand where the [(Squareroot of 6) / 2 (Square root of X)] comes from.

$\sqrt{6x}= \sqrt{6} \sqrt{x}$

so $\frac {d}{dx} \sqrt{6} \sqrt{x} = \sqrt{6} \ \frac {d}{dx} \sqrt{x} = \sqrt{6} \ \frac {1}{2} \frac {1}{\sqrt{x}}$

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