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Math Help - How does this work? (Derivatives)

  1. #1
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    How does this work? (Derivatives)

    I don't understand how the derivative of
    R(x)=\frac{1}{10000}(60000x-x^2)

    is

    R'(x)=\frac{1}{5000}(30000-x)


    I tried it with both the quotient rule and the product rule and I didn't get the same answer. Because isn't the derivative of a constant always 0?

    Please and thank you
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  2. #2
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    Think of R(x)=\frac{1}{10000}(60000x-x^2) as a constant times a variable.

    Apply the product rule:
    R(x)=\frac{1}{10000}(60000-2x) + 0(60000x-x^2)

    Combine terms:
    R(x)=\frac{60000-2x}{10000}

    Simplify:
    R(x)=\frac{30000-x}{5000} = \frac{1}{5000}(30000-x)
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  3. #3
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    Hello, CenturionMonkey!


    Are you simplifying your answers?

    You don't need any fancy rules.
    . . That fraction is a constant.


    R(x)\:=\:\frac{1}{10,000}(60,000x-x^2)

    We have: . R'(x) \;=\;\frac{1}{10,000}(60,000 - 2x) \;=\;\frac{2(30,000-x)}{10,000} \;=\;\frac{1}{5,000}(30,000-x)


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    If you insist on doing it the Long Way . . .


    Product Rule

    R(x) \;=\;\overbrace{\left(\frac{1}{10.000}\right)}^{f(  x)}\cdot\overbrace{(60,000x-x^2)}^{g(x)}

    R'(x)\;=\;\overbrace{\left(\frac{1}{10,000}\right)  }^{f(x)}\cdot\overbrace{(60000 - 2x)}^{g'(x)} + \overbrace{0}^{f'(x)}\cdot\overbrace{(60,000x - x^2)}^{g(x)} \;= \;\frac{1}{10,000}\cdot(60,000-2x)

    . . . . = \;\frac{1}{10,000}\cdot2(30,000 - x) \;=\;\frac{1}{5,000}(30,000-x)


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Quotient Rule

    R(x) \;=\;\frac{\overbrace{60,000x - x^2}^{f(x)}}{\underbrace{10,000}_{g(x)}}

    R'(x)\;=\;\frac{\overbrace{(10,000)}^{g(x)}\cdot\o  verbrace{(60,000-2x)}^{f'(x)} - \overbrace{(60,000x-x^2)}^{f(x)}\cdot\overbrace{0}^{g'(x)}}{\underbrac  e{10,000^2}_{[g(x)]^2}}

    . . . . = \;\frac{10,000(60,000 - 2x)}{10,000^2} \;=\;\frac{20,000(30,000-x)}{10,000^2} \;=\;\frac{30,000 - x}{5,000}

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  4. #4
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    Oh i see it now, thanks a lot guys. Great help
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