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anonymous

  • 5 years ago

2 + q^2/200 = 4000/p^2 differentiate implicitly

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  1. anonymous
    • 5 years ago
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    to find dq/dp

  2. anonymous
    • 5 years ago
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    yeh still not enough info , are we meant to assume that q ia a function of p

  3. anonymous
    • 5 years ago
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    sry...q is suppose demand for a good and p is the price of the good....

  4. anonymous
    • 5 years ago
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    yeh, so q is a function of p

  5. anonymous
    • 5 years ago
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    so it is the exact same as differentiating 2 + y^2 / 200 = 4000/x^2 where y=q=dependant variable

  6. anonymous
    • 5 years ago
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    \[2 + \frac{y^2}{200} = 4000x^{-2}\]

  7. anonymous
    • 5 years ago
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    diff wrt x ( and x=p ) 0 + (2y / 200 ) (dy/dx ) = -8000x^-3

  8. anonymous
    • 5 years ago
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    so dy/dx = -80000x^-3 / y

  9. anonymous
    • 5 years ago
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    \[\frac{dy}{dx} = \frac{-80000}{x^3 y}\]

  10. anonymous
    • 5 years ago
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    now replace y with q, and x with p

  11. anonymous
    • 5 years ago
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    ohhh ok thanks very much...

  12. anonymous
    • 5 years ago
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    wait, should be 800 000 on the top

  13. anonymous
    • 5 years ago
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    8000 x100 =800 000

  14. anonymous
    • 5 years ago
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    ok cool...i undersatnd your method though thats what i needed help with....

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spraguer (Moderator)
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