Let \(x\) and \(y\) by two quantities that do not depend on each other and are measured independently, say with measured values \(x_m\) and \(y_m\), and with measurement errors \({\vartriangle}x\) and \({\vartriangle}y\). Suppose you need a composite variable \(z=x\cdot y\). What will the error in \(z\) be? The linearization of \(z(x,y)\) plays a role in answering this question. After all, \[\begin{aligned}z(x_m+{\vartriangle}x,y_m+{\vartriangle}y) &\approx z(x_m,y_m)+z_x(x_m,y_m){\vartriangle}x+z_y(x_m,y_m){\vartriangle}y\\ &=z(x_m,y_m)+y_m{\vartriangle}x+x_m{\vartriangle}y\end{aligned}\] In other words: \[\frac{{\vartriangle}z}{z(x_m,y_m)}=\frac{{\vartriangle} x}{x_m} +\frac{{\vartriangle} y}{y_m}\] Because not the sign, but the size of the error is of interest, this suggests the following:

For multiplication, the relative error in the result is the sum of the relative errors in the factors.

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Let's have a look now at the quotient \(z(x,y)=\frac{x}{y}\) of variables. Via the total differential we find now the follwoing: \[\begin{aligned}{\vartriangle}z(x_m,y_m)&\approx z_x(x_m,y_m){\vartriangle}x+z_y(x_m,y_m){\vartriangle}y\\ &=\frac{{\vartriangle}x}{y_m}-\frac{x_m{\vartriangle}y}{y_m^2}\\ &=\frac{x_m}{y_m}\left(\frac{{\vartriangle}x}{x_m}-\frac{{\vartriangle}y}{y_m}\right)\end{aligned}\] ie \[\frac{{\vartriangle}z}{z(x_m,y_m)}=\frac{{\vartriangle} x}{x_m} -\frac{{\vartriangle} y}{y_m}\] Because not the sign, but the size of the error is of interest, this suggests the following:

For division. the relative error in the result is the sum of the relative errors in the numerator and denominator.

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So far we have noted that the total differential suggests the rules governing the propagation of error in formulas. This is because we do not take the sign of errors into account. If we do, then it makes sense to study squares of deviations. We look at \((\vartriangle{z})^2\) and we move in the above two cases to 'quadratic addition' of relative errors. This gives: