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Math Help - Product and Quotient Rule

  1. #1
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    Product and Quotient Rule

    In B.C, the sales of specialized high-endurance, out dollar gear in millions of dollars is projected by

    s(t)= 4t/t^2+2
    0<t<10

    where t is measured in years with t=0 corresponding to the beginning of 2000. How fast was the level of sales increasing at the beginning of 2005?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by lemontea View Post
    In B.C, the sales of specialized high-endurance, out dollar gear in millions of dollars is projected by

    s(t)= 4t/t^2+2
    0<t<10

    where t is measured in years with t=0 corresponding to the beginning of 2000. How fast was the level of sales increasing at the beginning of 2005?
    Do you mean : s(t)=\frac{4t}{t^2+2}?
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Do you mean : s(t)=\frac{4t}{t^2+2}?

    If this is the case, then find s'(5)

    Using Quotient Rule:

    s'(t)=\frac{(t^2+2)(4)-(4t)(2t)}{(t^2+2)^2}\implies s'(t)=\frac{-4t^2+8}{(t^2+2)^2}

    Thus, \color{red}\boxed{s'(5)=-\frac{92}{27^2}\approx -.1262}

    Using Product Rule:

    s(t)=(4t)(t^2+2)^{-1}

    \therefore s'(t)=4(t^2+2)^{-1}-(4t)(t^2+2)^{-2}\cdot (2t)
    \implies s'(t)=\frac{4}{t^2+2}-\frac{8t^2}{(t^2+2)^2}
    \implies s'(t)=\frac{4t^2+8-8t^2}{(t^2+2)^2}=\frac{-4t^2+8}{(t^2+2)^2}

    Thus, \color{red}\boxed{s'(5)=-\frac{92}{27^2}\approx -.1262}

    Hope this made sense!!
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