BELIEVE   ME   NOT!    - -     A   SKEPTICs   GUIDE  

Differentials

We have learned that the symbols  df  and  dx  represent the coupled changes in f(x) and x, in the limit where the change in x (and consequently also the change in f) become infinitesimally small. We call these symbols the differentials of f and x and distinguish them from $\Delta f$ and $\Delta x$ only in this sense: $\Delta f$ and $\Delta x$ can be any size, but df and dx are always infinitesimal - i.e. small enough so that we can treat f(x) as a straight line over an interval only dx wide.

This does not change the interpretation of the representation   ${\displaystyle {df \over dx}}$  for the derivative of f(x) with respect to x, but it allows us to think of these differentials df and dx as ``normal'' algebraic symbols that can be manipulated in the usual fashion. For instance, we can write

$$df = \left( df \over dx \right) dx$$

which looks rather trivial in this form. However, suppose we give the derivative its own name:

$$g(x) \equiv {df \over dx}$$

Then the previous equation reads

$$df = g(x) \; dx \qquad \hbox{\rm or just} \qquad df = g \; dx$$

which can now be read as an expression of the relationship between the two differentials df and dx. Hold that thought.

As an example, consider our familiar kinematical quantities

$$a \equiv {dv \over dt} \qquad \hbox{\rm and} \qquad v \equiv {dx \over dt}.$$

If we treat the differentials as simple algebraic symbols, we can invert the latter definition and write

$${1 \over v} = {dt \over dx}.$$

(Don't worry too much about what this ``means'' for now.) Then we can multiply the left side of the definition of a by 1/v and multiply the right side by dt/dx and get an equally valid equation:

$${a \over v} = {dv \over dt} \cdot {dt \over dx} = {dv \over dx}$$

or, multiplying both sides by  $v \, dx$,

$$a \; dx \; = \; v \; dv$$ (11.1)

which is a good example of a mathematical identity, in this case involving the differentials of distance and velocity. Hold that thought.