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Math Help - Derivatives and Tangents

  1. #1
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    Derivatives and Tangents

    Hey, there I have one quick question for now.. I'm in grade 12 calc and we're doing derivatives and have learned the power rule (its a nice shortcut) and i'm having trouble with this one problem:

    Write an equation of a tangent to each of the curves at the given point:

     y= \sqrt{3x^3} P(3,9)
    Heres what I did:

     y=(3x^3)^\frac{1}{2}
    Then the derivative:
    ...Do I use the power rule on the inside brackets first? or distribute the 1/2 outside the brackets??

     y'=(9x^2)^\frac{1}{2}
     y'=(3x)

    So my equation turns out to be
    9x-y-18=0,
    but the answer in the back of the book is 9x-2y-9=0
    What am I doing wrong?
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  2. #2
    Super Member
    Joined
    Dec 2008
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    Scotland
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    Quote Originally Posted by soulmeetsbody View Post
    Hey, there I have one quick question for now.. I'm in grade 12 calc and we're doing derivatives and have learned the power rule (its a nice shortcut) and i'm having trouble with this one problem:

    Write an equation of a tangent to each of the curves at the given point:

     y= \sqrt{3x^3} P(3,9)
    Heres what I did:

     y=(3x^3)^\frac{1}{2}
    Then the derivative:
    ...Do I use the power rule on the inside brackets first? or distribute the 1/2 outside the brackets??

     y'=(9x^2)^\frac{1}{2}
     y'=(3x)

    So my equation turns out to be
    9x-y-18=0,
    but the answer in the back of the book is 9x-2y-9=0
    What am I doing wrong?
    First of all, you should distribute the power!
     \frac{d}{dx} \sqrt{3x^3} = \frac{d}{dx} (3x^3)^{\frac{1}{2}}

      =  \frac{d}{dx}  (3^{\frac{1}{2}}x^{\frac{3}{2}})

      =  \frac{d}{dx}  (\sqrt{3}x^{\frac{3}{2}})

      = \sqrt{3} \frac{d}{dx}  x^{\frac{3}{2}}

      = \sqrt{3}  \times \frac{3}{2} \times  x^{\frac{1}{2}}

      = \frac{3\sqrt{3}}{2}  x^{\frac{1}{2}}

      =  \frac{3\sqrt{3}}{2}\sqrt{x}
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  3. #3
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    Jan 2009
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    Thank you very much! It makes a lot more sense now
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