Find the rate of change of change of x with respect to p:
$\displaystyle p=\sqrt \frac {500-x}{2x}$ , $\displaystyle 0<x<500$
If you could guide me through this problem that would be great.
rate of change can be viewed as slope geometrically. now certainly you want the instantaneous rate of change of x with respect to p.
ie $\displaystyle \frac{dx}{dp}$ known as the derivative of x with respect to p.
so let's find the slope. remember that the p is shown respect here so it is our variable.
im not sure how far into calc you are but i'll do it the long way
$\displaystyle p^2=\frac{500-x}{2x}$
$\displaystyle 2p^2x+x=500$
$\displaystyle x(2p^2+1)=500$
$\displaystyle x=\frac{500}{2p^2+1}$
now we take the derivative
inst. rate of change$\displaystyle =x'=\frac{dx}{dp}=\frac{d}{dp}\left[\frac{500}{2p^2+1}\right]=\frac{(2p^2+1)0-500(4p)}{(2p^2+1)^2}$