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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
 one year ago
 one year 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
 one year ago
 one year ago

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oioBest ResponseYou've already chosen the best response.0
\[y=f(p(t))10\sqrt{810000\left[ p \left( t \right) \right]^{2}}\] \[p(t)=\frac{ 400 }{ 1+\sqrt{t} }+500\] I hope this is more helpful :(
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
I don't understand what the question wants. Two equations and that's it?
 one year ago

oioBest ResponseYou've already chosen the best response.0
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!
 one year ago

oioBest ResponseYou've already chosen the best response.0
Thanks guys for helping me! :)
 one year ago

oioBest ResponseYou've already chosen the best response.0
Thank you for replying at the moment
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
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
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
p′(t)=(400(1+t√)−1+500)' sorry mistype
 one year ago

oioBest ResponseYou've already chosen the best response.0
thank you but...im still kind of confused... :( I really get confused when I derive f(p(t)) itself
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
yup, you might want to let it as u before starting...
 one year ago
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