The demand $x$ is the number of items that can be sold at a"price of $\mathrm{\$}p$. For $x=\sqrt[3]{6,000-p^3}$, find the rate of change of $p$ with respect to $x$ by differentiating implicitly.

Step 1 ：Given the equation $x=\sqrt[3]{6000-p^3}$, we want to find the rate of change of $p$ with respect to $x$. This is equivalent to finding $\frac{dp}{dx}$.

Step 2 ：To do this, we first need to differentiate both sides of the equation with respect to $x$.

Step 3 ：Starting with the left side, the derivative of $x$ with respect to $x$ is simply 1.

Step 4 ：For the right side, we use the chain rule for differentiation. The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function.

Step 5 ：The outer function is $\sqrt[3]{u}$ and the inner function is $6000-p^3$. The derivative of the outer function is $\frac{1}{3}{u}^{-2/3}$ and the derivative of the inner function is $-3p^2$.

Step 6 ：Applying the chain rule, we get $\frac{1}{3}(6000-p^3{)}^{-2/3}\ast (-3p^2)$.

Step 7 ：Simplifying this expression, we get $-p^2(6000-p^3{)}^{-2/3}$.

Step 8 ：Setting the derivatives equal to each other, we get $1=-p^2(6000-p^3{)}^{-2/3}$.

Step 9 ：Solving for $\frac{dp}{dx}$, we get $\frac{dp}{dx}=-\frac{1}{p^2(6000-p^3{)}^{2/3}}$.

Step 10 ：Thus, the rate of change of $p$ with respect to $x$ is $\boxed{{\displaystyle -\frac{1}{p^2(6000-p^3{)}^{2/3}}}}$.