Can dy/dx Be Treated as a Fraction?
Date: 08/26/2004 at 14:57:25 From: Amit Subject: Properties of dy/dx When I learned about derivatives, I learned that dy/dx was a notation that implied "derivative of y with respect to x." I understood that. But I am confused about whether or not the notation dy/dx can be treated as a fraction, giving individual meanings to dy and dx. For example, in integration by substitution: integrate (sin(3x+5)dx) u = 3x+5 du/dx = 3 dx = du/3 That is the part that confuses me. How can the dx be solved for? What exactly is "dx"? I understand that dy/dx is a limit, and that it is a slope. But the idea of it being simply a notation doesn't help me understand how you can multiply out the bottom. Any help would be appreciated...
Date: 08/26/2004 at 15:27:40 From: Doctor Vogler Subject: Re: Properties of dy/dx Hi Amit, Thanks for writing to Dr Math. The easy answer to your question is that your definition for dy/dx is correct; it means the derivative of y with respect to x, and dy and dx are meaningless when written alone, so that dx = du/3 is not a meaningful expression but should be written dx/du = 1/3. And when certain nice things happen that *look* like fractions, such as: 1 dy/dz = ----- dz/dy and dz/dx = dz/dy * dy/dx then this is actually just the Chain Rule at work. And the reason that integral( f(g(x)) g'(x) dx ) = integral( f(u) du ) is not that u = g(x) implies du = g'(x) dx but rather the Chain Rule again. All of that is true, except that I should qualify the "not a meaningful expression." You see, something is only meaningless until somebody gives it a formal meaning. Then you hope that the meaning they gave it has useful properties (such as, that it relates to derivatives...). In fact, this has been done, and there is a good deal of mathematics that has gone into the theory of differentials, and it fits into integrals, and putting the differential "dx" at the end of every integral also makes sense according to this theory, and so on. One math doctor alluded to some of this on Differentials http://mathforum.org/library/drmath/view/53678.html You can also get books that discuss this in more detail. But the fact is that most people who use calculus don't really need all of the theory of differentials, and the Chain Rule indeed suffices to verify most facts that you would get from treating dy/dx as a fraction. The reason you can treat it as a fraction is that dy/dx is the limit of a fraction, and so most of the operations you would do to the fraction you can do before you take the limit. In other words, *before* the limit is taken, it *is* a fraction, so you can treat it as one. But then you take the limit and it becomes a derivative. Finally, there is also the theory of estimating with derivatives, where I always say to think of dx = change in x dy = change in y x = unchanged value of x y = unchanged value of y For example, to estimate (1.98)^6, we use y = x^6 dy = 6*x^5 dx x = 2 dx = -0.02 (so that x + dx = 1.98), and therefore y = 2^6 = 64 dy = 6*x^5 dx = 6*32*-0.02 = -3.84 which implies that y + dy = 64 - 3.84 = 60.16, which is, in fact, a pretty close approximation to (1.98)^6. And this is essentially treating dy/dx as a fraction before we've taken the limit, since dx doesn't go all the way to zero but only to -0.02. Does this help you to understand how differentials are a fraction in some sense but not in others? If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum