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Suppose the ideal gas equation $R=\frac{PV}{T}$ has $R$ as constant. If the percentage error in measuring $P,V,T$ are $1$%, $2$% and $3$%, how to find the maximum percentage error in $R$? In this question, I think we first need to find the differential of $R$ $$dR=\frac{V}{T}dP+\frac{P}{T}dV-\frac{PV}{T^2}dT$$ Dividing both sides by $R$, we get $$\frac{dR}{R}=\frac{dP}{P}+\frac{dV}{V}-\frac{dT}{T}$$

But since we do not the know the exact value of $V, T$ and $P$, how are we going to find the result?

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    Hint: Divide both sides of the above equation by $R$. – 2017-02-04
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    Sorry, I still couldn't derive the result. – 2017-02-04
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    Go to $\Delta$'s. Everything will correspond to relative errors. – 2017-02-04
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    I would straightforward use the numbers. The numerator has to be as large as possible and the denominator has to be as small as possible. $max \ dR/R=\frac{1.01\cdot 1.02}{0.97}-1$ – 2017-02-04
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    @callculus: As the error can go in both sides, you need also to compute $\frac{0.99·0.98}{1.03}-1$ to capture the full range of possible values. – 2017-02-04
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    @LutzL Yes if it is asked for the range. But I thought that it is asked for the maximum difference of the true value and the measured value. I get $6.2\%$ – 2017-02-04
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    @callculus That is true, but one still has to lose a word or two on why the second variant gives less an absolute error than the first. – 2017-02-04
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    @LutzL You´re right, I forgot that. – 2017-02-04

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The percentage error is the error divided by the true value, which is exactly what you have. You are given $\left|\frac {\Delta V}V\right| \lt 0.02$ and so on. You are being asked to compute the upper limit on $\left|\frac {\Delta R}R\right| $