We have a quantity $a$ expressed in terms of two quantities $b $ and $c$ as $a = b/c$. It seems to me that there are two ways of estimating the error on $a$, the "physics" prescription and a "maths" one based on differentiation:

  1. According to "physics": $∆a/a = ∆b/b + ∆c/c$

  2. According to "math" (namely, using differentiation): $da/a = db/b - dc/c$

Which is wrong?

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    $\begingroup$ Both are wrong. When you learn more, you will get to see the correct derivation that will agree between physics and maths. For now, take the physics version, because you do not yet have to correct maths for it. $\endgroup$ Jul 20, 2023 at 16:35
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    $\begingroup$ I've definitely not seen that proclaimed formula in physics anywhere. I've taught both calculus and non-calculus based physics labs where students are required to do proper error analysis and that formula wasn't listed in the non-calculus table (since they weren't required to know how the formula were derived). $\endgroup$
    – Triatticus
    Jul 20, 2023 at 16:59
  • $\begingroup$ @naturallyInconsistent okay just tell me correct expression $\endgroup$ Jul 22, 2023 at 1:56
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    $\begingroup$ You have to use variances, which implies squaring and other stuff that will get rid of minus signs. The proper derivation using partial derivatives and variance and considering the covariances are everywhere on the internet. The search term is propagation of uncertainty. $\endgroup$ Jul 22, 2023 at 5:53

2 Answers 2


Consider two positive quantities $b>0$ and $c>0$ with associated "small errors" $\delta b>0$ and $\delta c>0$, and

$$a(x) = \dfrac{b+x \delta b}{c-x \delta c}$$

where $x\in(-1,1)$ is an extra parameter that works as a sort of "bookmark": $x$ keeps track of the "small" elements in the expression, namely $\delta b$ and $\delta c$. In which range do we expect $a(x)$ to vary? The minimum is for $x=-1$ (minimize the numerator and maximize the denominator!) and the maximum is for $x=1$: $$ a_m = \dfrac{b- \delta b}{c+ \delta c} < a(x) < \dfrac{b+ \delta b}{c- \delta c}=a_M $$

To the linear order in $x$ (we can do this despite the fact that $x$ is not necessarily small because $x$ bookmarks the small errors!), we have $$ a(x) \approx \dfrac{b}{c}+x \dfrac{c \, \delta b + b \, \delta c}{c^2}+ O(x^2) $$ In the above expression, $O(x^2)$ means that we stop the expansion before quantities like $(\delta b)^2$, $(\delta c)^2$, $ \delta b \,\delta c $ appear. In this approximation, $a_m$ and $a_M$ still correspond to $x=- 1$ and $x=1$, so that $$ a_m \approx \dfrac{b}{c}- \dfrac{c \, \delta b + b \, \delta c}{c^2} \qquad \qquad a_M \approx \dfrac{b}{c}+ \dfrac{c \, \delta b + b \, \delta c}{c^2} $$ In the end, we obtain the expression for $\delta a>0$ by considering the half-distance between $a_m$ and $a_M$: $$ \delta a = \dfrac{a_M-a_m}{2}\approx \dfrac{c \, \delta b + b \, \delta c}{c^2} = \dfrac{ \delta b}{c} + a \dfrac{\delta c}{c} \quad \Rightarrow \quad \dfrac{\delta a}{a} \approx \dfrac{ \delta b}{b} + \dfrac{\delta c}{c} $$

Extra: You may use the same method to justify the error propagation for the product $a=bc$ by considering $$a(x)=(b+x \delta b)(c+x\delta c)$$ for $0<\delta b <b$ and $0<\delta c <c$. Again, $a_m=a(-1)$ and $a_M =a(1)$ and $\delta a =(a_M-a_m)/2$. The formula to the linear order in $x$ is the same as the one for the quotient discussed above, see this.

Important note: The theory of propagation of uncertainty and the (more advanced) uncertainty quantification provide a better (statistical) justification for this simple result, as well as its validity range.

"Quadratic" approach (see the comments) - Given a random variable $X$ that is a linear combination of other two random variables, e.g. $X=sY+rZ$, we have that (see this for the proof or Wikipedia):

$$ Var(X) = s^2Var (Y) + r^2Var(Z) + 2sr Cov(Y,Z) $$

where the covariance is: ${Cov}(Z,Y) = {E}\left[ (Z-{E}[Z]) (Y-{E}[Y]) \right]$.

Now, if $X=f(Y,Z)$, we can use the above formula if we take the linear approximation $dX \approx \partial_Y f dY + \partial_Z f dZ $: we have that $s=\partial_Y f$ and $r=\partial_Z f$. The deviations $dX$, $dY$, $dZ$ are legit random variables: $dX= X-f(E(Y),E(Z))$, $dY=Y-E(Y)$, $dZ=Z-E(Z)$, where $E$ denotes the expected value. Moreover, $Var(X+c)=Var(X)$ for any constant $c$, so that $Var(X)=Var(dX)$. the same is valid for $Y,Z$.

In the end:

$$ Var(X) = Var(dX) \approx (\partial_Y f)^2 Var(Y)+ (\partial_Z f)^2 Var(Z) + 2 (\partial_Y f)( \partial_Z f) Cov(Y,Z) $$

We can apply this formula to $a=bc$ (the case $a=b/c$ is analogous): $c$ plays the role of the random variable $X$ and the values $a,b$ are realizations of the random variables $Y,Z$. We also assume that $Y,Z$ are independent (so their covariance is zero). Therefore, $f=bc$, $\partial_bf=c$, $\partial_cf=b$ (for consistency, these derivatives are formally calculated at the observed/expected value) and

$$ Var(a) \approx b^2 Var(c)+ c^2 Var(b) \qquad \Rightarrow \qquad \dfrac{Var(a)}{a^2} \approx \dfrac{Var(c)}{c^2} + \dfrac{Var(b)}{b^2} $$

You may be surprised to see that the result is quadratic, but this is not mysterious: Why is propagation of uncertainties quadratic rather than linear?. As said, the simple prescription discussed in the first part of this answer is just a convenient approximate way of estimating the expected interval in which the variable $a$ can vary. Better estimation of the variance of $X$ can be given by considering the algebra of random variables.

  • $\begingroup$ I got it that how this formula can be derived but also tell me that what's wrong with differentiation method? $\endgroup$ Jul 22, 2023 at 1:55
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    $\begingroup$ This is a quick and dirty derivation with the "differentiation method"! You just consider the 1-parameter curve $a(x)$ because you want to know "in which interval $a$ can vary". In both questions, me and @Farcher are saying the same thing. Please note that, to fully understand "uncertainty propagation", you have to start from considering variances or absolute deviations, but this requires more advanced math (if you want I can expand the answer). Anyway, the correct prescription is what you call "physics" prescription. Here I show you how to obtain it by using "differentiation". $\endgroup$
    – Quillo
    Jul 22, 2023 at 6:34
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    $\begingroup$ @GovindPrajapat I would appreciate it if you could be more specific in your request: the answer could be expanded in a series of ways, what is the exact point that is not clear? :) $\endgroup$
    – Quillo
    Jul 25, 2023 at 15:58
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    $\begingroup$ Done! Hope that helps. $\endgroup$
    – Quillo
    Jul 25, 2023 at 17:35
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    $\begingroup$ @GovindPrajapat the plus/minus sign is used to give you the idea that you have to follow the procedure at the beginning of the answer: $a\pm \delta a \in [a-\delta a , a+ \delta a ]$. It just notation. Also, do not forget to accept an answer that somehow clarified your initial doubts. $\endgroup$
    – Quillo
    Jul 26, 2023 at 12:15

There is nothing wrong with the equation $\dfrac {\delta a}{a} = \dfrac {\delta b}{b} - \dfrac {\delta c}{c} $ provided it is used correctly and it is equivalent to $\dfrac {\Delta a}{a} = \dfrac {\Delta b}{b} + \dfrac {\Delta c}{c} $.

At first sight the equation $\dfrac {\delta a}{a} = \dfrac {\delta b}{b} - \dfrac {\delta c}{c} $ seems all wrong for calculating errors particularly if the fractional changes in $b$ and $c$ are equal as that gives a fractional change in $a$ of zero.

Does that really mean that the fractional error in $a$ is zero?
The answer is yes, that could be true.

But what are you actually looking for?
With simple error analysis one is looking for the maximum possible fractional error in $a$ given fractional errors in $b$ and $c$.

If $a=\dfrac bc$ how does one compute the maximum value of $a$?

It is done by evaluating $\dfrac{\text{maximum value of b}}{\text{minimum value of c }}= \dfrac {b+\delta b}{c-\delta c}$.

Similarly the minimum value of $a$ would be $\dfrac {b-\delta b}{c+\delta c}$.

So if one were to use the equation $\dfrac {\delta a}{a} = \dfrac {\delta b}{b} - \dfrac {\delta c}{c} $ one would have to assign the appropriate signs to $\delta b$ and $\delta c$ to maximize (or minimise) the fractional change in $a$.
Thus for a maximum, $\delta b$ would be positive and $\delta c$ would be negative.

Rather than go through all this palaver, for error analysis the equation $\dfrac {\Delta a}{a} = \dfrac {\Delta b}{b} + \dfrac {\Delta c}{c} $ is used with $\Delta b$ and $\Delta c$ both being positive which immediately gives the maximum fractional error in $a$.

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    $\begingroup$ Why we have to maximize or minimise the fractional change in a by taking δb positive and δc negative. $\endgroup$ Jul 21, 2023 at 2:39
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    $\begingroup$ This method makes a rough estimate of the maximum possible error. There are better estimators of the error but this method has the advantage of simplicity in terms of explanation and execution. $\endgroup$
    – Farcher
    Jul 21, 2023 at 6:26
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    $\begingroup$ @GovindPrajapat that's the important point: you are trying to estimate (estimate!) in which interval the quantity $a=b/c$ can vary by finding its maximum and minimum. The same result can be obtained (a bit more rigorously) by considering the "sum in quadrature" of the uncertainties, see physics.stackexchange.com/q/233724/226902. Even this is, however, a result that is limited in scope (it assumes normally distributed variables). In the end, this is a simple prescription that kind of works and that can be somehow justified in different ways. $\endgroup$
    – Quillo
    Jul 21, 2023 at 9:18
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    $\begingroup$ @Quillo: Nitpick: Normally and independently distributed variables. In fact, in the extreme case the quadrature formula actually reduces to the linear sum of the fractional uncertainties. $\endgroup$ Jul 21, 2023 at 11:39
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    $\begingroup$ @MichaelSeifert thank you for the important remark. This point is also widely explained in many linked PSE posts: it is important not to give the impression that what discussed in this question constitutes a real "proof". More on this: Combining uncertainties - multiple measurements. $\endgroup$
    – Quillo
    Jul 21, 2023 at 13:11

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