My book deals with the combination of errors for addition, multiplication and for exponents. I have understood the derivation of each but I am struggling to extend the same for division and negative powers.

For Errors of Multiplication : Let $Z$ = $AB$, thus $Z \pm \delta Z$ = $(A\pm\delta A)(B\pm \delta B)$

i.e : $Z \pm \delta Z$ = $AB\pm B\delta A\pm A\delta B\pm \delta A\delta B$

Dividing $LHS$ by $Z$ and $RHS$ by $AB$ gives :

$\frac{\delta Z}{Z}$ = $\frac {\delta A}{A}$ $\pm$ $\frac{\delta B}{B}$

Question 1:

$a)$ For the derivation regarding division how do you proceed beyond $Z \pm \delta Z$ = $\frac{$(A\pm\delta A)}{(B\pm \delta B)$}$?

$b)$ If not via the above method , how is the result for division derived?

For Errors in Exponents : $Z$ = $A^2$ which gives : $\frac{\delta Z}{Z}$ = $\frac{$2\delta A$}{A}$

This is generalized to give : $\frac{$\delta Z$}{Z}$ = $p \frac{\delta A}{A}$+ $q\frac{\delta B}{B}$+$r\frac{\delta C}{C}$ where $Z$ = $\frac{(A^p)(B^q)}{C^r}$

Question 2:

$c)$ How is the result arrived at for negative powers?

  • $\begingroup$ I forgot to precise that $\frac{\Delta X}X$ stands for $\left|\frac{\Delta X}{X}\right|$ $\endgroup$ Commented Apr 7, 2019 at 9:10

1 Answer 1


In my humble opinion, the most convenient way is to use logarithms and logarithmic differnetiation. $$Z=\frac A b\implies \log(Z)=\log(A)-\log(B)$$ making $$\frac{\Delta Z}Z=\frac{\Delta A}A+\frac{\Delta B}B$$ since the errors add. You will for sure notice that this is the same as for $Z=A\,B$ .

You can do the same for $Z=A^a \,B^b\, C^c$ where the exponents can be positive or negative (or even $0$) to get $$\frac{\Delta Z}Z=|a|\frac{\Delta A}A+|b|\frac{\Delta B}B+|c|\frac{\Delta C}C$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .