The radius of a sphere is measured as 5 cm ± 0·1 cm. Use differentiation to find the volume of the sphere in the form Vcm^3 ± bcm^3. I have tried:
V = 4/3πr^3
If the radius is 5 ± 0·1, then V = 4/3π(5 ± 0·1) ^3
giving V = 555.65 or V = 492.81
The provided solution is (524 ± 31) cm^3 but I am not sure how you derive this result. This is from a Year 12 Maths Methods textbook.

This is a "linearization" or "approximation by differentials" problem.  The goal is to approximate the error in calculating the volume, knowing the error in measuring the radius.  Since $V=\frac{4}{3}\pi r^3$, then $V'= 4 \pi r^2$, and  you have:
$$\Delta V \approx dV = 4 \pi r^2 \; dr$$
Let the error in the radius be $\Delta r = dr = 0.1$ and $r=5$ to get
$$\Delta V \approx 4 \pi (5)^2 (0.1)= 31.4\ldots.$$
