Question

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

Answer 1

@dpaInc

Answer 2

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

Answer 3

I don't understand what the question wants. Two equations and that's it?

Answer 4

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!

Answer 5

Thanks guys for helping me! :)

Answer 6

Thank you for replying at the moment

Answer 7

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

Answer 8

p′(t)=(400(1+t√)−1+500)' sorry mistype

Answer 9

thank you but...im still kind of confused... :( I really get confused when I derive f(p(t)) itself

Answer 10

yup, you might want to let it as u before starting...

Answer 11

uhm thanks :)