Rate of change - Implicit differentiation A price $p$ (in dollars) and demand $x$ for a product are related by $$\left(2x^2\right)-2xp+50p^2 = 20600$$ If the price is increasing at a rate of $2$ dollars per month when the price is $20$ dollars, find the rate of change of the demand.


I was a little confused on how to proceed with this question. Am I supposed to use implicit differentiation (with the $x$ serving the same purpose as a $y$) and then find the derivative of $x$?

This is the implicit differentiation I tried:

$$4x\frac{dx}{dp}-2\frac{dx}{dp}+100p = 0$$

$$4x\frac{dx}{dp}-2\frac{dx}{dp} = -100p$$

$$\frac{dx}{dp}(4x-2) = -100p$$

$$\frac{dx}{dp} = \frac{-100p}{4x-2}$$

I believe this is the derivative I am looking for (thought not entirely sure) but I am not sure what values of $p$ and $x$ to input, as I am supposed to get a numerical final answer.

Any help?
Guide:

$$2x^2-2xp+50p^2=20600$$

Differentiate with respect to $t$.

We have information about $x, p, \frac{dp}{dt}$, and you are interested in finding $\frac{dx}{dt}$.

Note that when $p=20$, you can compute the corresponding $x$.
Blindly follow the rule that $\boxed{{\rm d}f = \frac{\partial f}{\partial x} {\rm d}x + \frac{\partial f}{\partial p} {\rm d}p}$.

In this case:

$$ 2x^2-2xp+50p^2 = 20600 $$

$$ {\rm d} \left( 2x^2-2xp+50p^2 = 20600  \right)  $$

$$ \frac{\partial}{\partial x} \left( 2x^2-2xp+50p^2 = 20600  \right) {\rm d}x + \frac{\partial}{\partial p} \left( 2x^2-2xp+50p^2 = 20600  \right) {\rm d}p$$

$$ \left( 4x -2p  \right) {\rm d}x + \left(100p - 2x  \right) {\rm d} p = 0 $$

$$ \frac{{\rm d}x}{{\rm d}p} = \frac{x-50 p}{2 x -p} $$