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Math Help - Calculate Derivative? Square root?

  1. #1
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    Question Calculate Derivative? Square root?


    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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  2. #2
    Super Member Random Variable's Avatar
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    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}
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  3. #3
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    Question

    I still don't understand where the [(Squareroot of 6) / 2 (Square root of X)] comes from.
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  4. #4
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Brazuca View Post
    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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