# Uncertainty Calculation: Applying Product Rule instead of Power Rule

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?

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.