### Differentials and integrals: Differentials

### Application: error analysis

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):

*In the calculation of a function value \(f(x_m)\) of a measured result \(x_m\), the inaccuracy in \(x_m\) is multiplied by the absolute value of the derivative \(f'(x_m)\).*