The volume $V$ of a sphere with diameter $D$ is given by the formula $V=\tfrac{1}{6}\pi D^3$. Suppose that the diameter of a sphere is determined with a vernier caliper and assigned to the variable $D_\mathrm{m}$ and that the measurement error is equal to ${\vartriangle}D$, in other words, the true value for the diameter lies with great reliability between $D_\mathrm{m}-{\vartriangle}D$ and $D_\mathrm{m}+{\vartriangle}D$. This leads to the next deviation ${\vartriangle}V$ of the true volume:

$$\begin{aligned} {\vartriangle}V &=\tfrac{1}{6}\!\pi\left((D_\mathrm{m}+{\vartriangle}D)^3-D_{m}^3\right)\\ &= \tfrac{1}{6}\!\pi\left(3D_\mathrm{m}^2\cdot{\vartriangle}D+3D_\mathrm{m}\cdot({\vartriangle}D)^2+ ({\vartriangle}D_\mathrm{m})^3\right)\\ &\approx \tfrac{1}{2}\!\pi D_\mathrm{m}^2\cdot {\vartriangle}D, \text{ for small }{\vartriangle}D.\\ \end{aligned}$$ Thus, the error in the determination of the volume of the sphere can be estimated as $V'(D_\mathrm{m})\cdot{\vartriangle}D$ .

More generally (but in somewhat sloppy wording):