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I use $\delta$ to represent absolute uncertainty. The power rule for the calculation of relative uncertainty in $t^2$ is $$\frac{\delta (t^2)}{(t^2)}=2\left(\frac{\delta t}{t}\right).$$ But if I treat the power as a product and apply the product rule, I have $$\frac{\delta (t \times t)}{(t \times t)} = \sqrt{\left(\frac{\delta t}{t}\right)^2 + \left(\frac{\delta t}{t}\right)^2} = \sqrt{2\left(\frac{\delta t}{t}\right)^2} = \sqrt{2}\left(\frac{\delta t}{t}\right).$$ Am I making a mistake? If not, how is this inconsistency reconciled?

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The product rule assumes that the things being multiplied vary independently of one another, which is clearly not the case when multiplying something by itself. As such, the power rule is the correct one here.

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  • $\begingroup$ Thank you. In what part of the derivation of these rules is this assumption necessitated? Or is it not and we just find that these rules are only consistent with independent variables? [I'm staring at the propagation of error formula with partial derivatives.] $\endgroup$ Mar 1, 2023 at 1:57
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    $\begingroup$ @BoundaryCondition Adding uncertainties in quadrature (square root of the sum of squares) is the specific step that assumes independence. $\endgroup$
    – Sandejo
    Mar 1, 2023 at 2:05
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As was pointed out by Sandejo's answer, uncertainties only add in quadrature if the two quantities being measured are uncorrelated. The formulas for propagation of uncertainty have additional terms that must be included if the variables' uncertainties are correlated.

Specifically, suppose we have two quantities $A$ and $B$ with uncertainties $\sigma_A$ and $\sigma_B$ and covariance $\sigma_{AB}$. Then the uncertainty in $f = AB$ is [given by] $$ \frac{\sigma_f}{f} = \sqrt{\frac{\sigma_A^2}{A^2} + \frac{\sigma_B^2}{B^2} + 2 \frac{\sigma_{AB}}{AB}}. $$ Uncorrelated uncertainties have $\sigma_{AB} = 0$, and so you end up with the "addition in quadrature" formula you're familiar with.

On the other hand, suppose that $A$ and $B$ are perfectly correlated (as they would be if they were secretly the same variable, which you're calling $t$.) In this case, it can be shown that $\sigma_{AB} = \sigma_A \sigma_B$, and so $$ \frac{\sigma_f}{f} = \sqrt{\frac{\sigma_A^2}{A^2} + \frac{\sigma_B^2}{B^2} + 2 \frac{\sigma_{A}\sigma_{B}}{AB}} = \frac{\sigma_A}{A} + \frac{\sigma_B}{B}. $$ In particular, if $A = B = t$, then you have $$ \frac{\sigma_f}{f} = 2 \frac{\sigma_t}{t}, $$ as one would expect from the power rule.

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