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Thread: 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:

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

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

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

    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
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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:

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

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

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

    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!
    $\displaystyle \frac{d}{dx} \sqrt{3x^3} = \frac{d}{dx} (3x^3)^{\frac{1}{2}} $

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

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

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

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

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

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