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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/ 
Associated Topics:
College Calculus
High School Calculus

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