# Treating a differential as a difference of two quantities---legal or no? [duplicate]

In some textbooks, I have seen the authors treat differentials such as $dx$ as a difference, so $dx$ = $x_2$ - $x_1$

My question is, how legal is this? I realize the differential is a quantity smaller than any conceivable real number but shouldn't this be $dx$ $\approx$ $x_2$ - $x_1$?

• I voted to migrate this to Math.SE. – AccidentalFourierTransform Jun 5 '18 at 18:35
• The limit of a difference is actually the more rigorous way to think about it, because that's how the differential is defined in the definition of the derivative. An isolated differential doesn't mean anything without another differential on the opposite side of the equation; in that situation, the equation represents a linear approximation for how a difference in one quantity induces a difference in another. – probably_someone Jun 5 '18 at 18:41
• I posted this for physicists because I wanted their take on the use of differentials. Mathematicians tend to be put off by the non-rigorous use of mathematics that physicists work with. – Steven Jun 5 '18 at 19:32
• @Steven - That's true. Physicists tend to be more careless than mathematicians with mathematical notations and concepts because they're only interested in using the mathematical tools to get the right answer, even if the procedure that they use may be considered to be sloppy by mathematicians. So, yes, the procedure that the authors of your textbook used may not be strictly "legal" from a mathematician's standpoint. Can't say anything more without knowing exactly how the authors of your textbook used the differentials. – user93237 Jun 5 '18 at 19:57
• @SamuelWeir Here is an example taken from S.A. Elder's book "Fluid Physics for Oceanographers and Physicists": "A dry column of the atmosphere has a constant lapse rate from the surface to an altitude of 2000 meters. If the sea level temperature is 0 degrees C, and the temperature at 2000 meters is 30 degrees C, is this air column stable?" He then proceeds to calculate the lapse rate as $\lambda$ = $\frac{dT}{dh}$ = -(-30-0)/(2000-0) – Steven Jun 5 '18 at 22:54