Let $r$ be the radius and $x$ be the diameter of the sphere. We know that $x = 2r$, which implies $r = \frac{x}{2}$.
The volume $V$ of a sphere is given by $V = \frac{4}{3}\pi r^3$. Substituting $r = \frac{x}{2}$ into the formula: $$V = \frac{4}{3}\pi \left(\frac{x}{2}\right)^3 = \frac{4}{3}\pi \left(\frac{x^3}{8}\right) = \frac{\pi x^3}{6}$$
We need to find the rate of change of volume with respect to diameter, which is $\frac{dV}{dx}$: $$\frac{dV}{dx} = \frac{d}{dx} \left(\frac{\pi x^3}{6}\right) = \frac{\pi}{6} \cdot 3x^2 = \frac{\pi x^2}{2}$$
Given $r = 5$ cm, the diameter $x = 2r = 10$ cm. Substituting $x = 10$ into the derivative: $$\frac{dV}{dx} = \frac{\pi (10)^2}{2} = \frac{100\pi}{2} = 50\pi$$
Final Answer: 50\pi~cm^{3}/cm
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