anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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saifoo.khan
  • saifoo.khan
@dpaInc
anonymous
  • anonymous
\[y=f(p(t))10\sqrt{810000-\left[ p \left( t \right) \right]^{2}}\] \[p(t)=\frac{ 400 }{ 1+\sqrt{t} }+500\] I hope this is more helpful :(
anonymous
  • anonymous
I don't understand what the question wants. Two equations and that's it?

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anonymous
  • anonymous
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!
anonymous
  • anonymous
Thanks guys for helping me! :)
anonymous
  • anonymous
Thank you for replying at the moment
anonymous
  • anonymous
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
anonymous
  • anonymous
p′(t)=(400(1+t√)−1+500)' sorry mistype
anonymous
  • anonymous
thank you but...im still kind of confused... :( I really get confused when I derive f(p(t)) itself
anonymous
  • anonymous
yup, you might want to let it as u before starting...
anonymous
  • anonymous
uhm thanks :)

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