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if W=1/X is the inverse of X. show that W=w±Δw with w=1/x and Δw/w= Δx/x. Thus, the percentage error of the inverse of a quantity is the same as that of the quantity.
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\[\ln W=-\ln x\] So, \[\Delta\ln W=-\Delta\ln x\] i.e. \[\frac{\Delta W}{W}=-\frac{\Delta x}{x}\]
Another way is Wx=1, Thus, \[\Delta (Wx)=0\] but then, \[\Delta (Wx)=W\Delta x+x\Delta W=0\] which again means that, \[\frac{\Delta W}{W}=-\frac{\Delta x}{x}\]
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