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oio

  • 3 years ago

Hi! This question deals with derivatives Ok, here it is: The quantity demanded per month, y, of USEM tablets is related to the average price unit, p(t) (in dollars), of tablets by the equation y=f(p(t))= 10 * sqrt(810,000-[p(t)]^2) It is estimated that 3 months from now, the average price of the tablet will be given by p(t)= [400/(1+sqrt(t))] +500 (t is less than or equal to 60 and more than or equal to 0) I know I should use the Product and The Chain Rule, but I'm not sure how to apply it! Thank you in advance. I just need to find the derivative! I would really love an exp

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  1. saifoo.khan
    • 3 years ago
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    @dpaInc

  2. oio
    • 3 years ago
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    \[y=f(p(t))10\sqrt{810000-\left[ p \left( t \right) \right]^{2}}\] \[p(t)=\frac{ 400 }{ 1+\sqrt{t} }+500\] I hope this is more helpful :(

  3. Shadowys
    • 3 years ago
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    I don't understand what the question wants. Two equations and that's it?

  4. oio
    • 3 years ago
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    oh whooop! AUGGHH! i forgot the rest hang on :) Find the rate at which the quantity of tablets deamnded per month will be changing 16 months from now!

  5. oio
    • 3 years ago
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    Thanks guys for helping me! :)

  6. oio
    • 3 years ago
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    Thank you for replying at the moment

  7. Shadowys
    • 3 years ago
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    LOL sorry. You might notice that \[f(p(t))' = f'(p(t))* p'(t)\] And you have both. Now, for p(t), \[p'(t)=400(1+\sqrt t)^-1 +500\] To use the chain rule, you can let u =(1+\sqrt t)^-1

  8. Shadowys
    • 3 years ago
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    p′(t)=(400(1+t√)−1+500)' sorry mistype

  9. oio
    • 3 years ago
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    thank you but...im still kind of confused... :( I really get confused when I derive f(p(t)) itself

  10. Shadowys
    • 3 years ago
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    yup, you might want to let it as u before starting...

  11. oio
    • 3 years ago
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    uhm thanks :)

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