~basic idea: Multiplying and/or dividing certain derivatives in order
                      to obtain a desired derivative.

~there are several forms of the chain rule.

1)   Multiplying derivatives:  
      
      This form is widely used to expand our formula base.
      It is sometimes called the composite form, since it is used for    
      derivatives of functions "within" functions. See the calculus
      web site link on my home page for a very nice discussion.
     ~Note: This is its form: If h(x)=g[f(x)], then h ' (x)=g '[f(x)] f ' (x)

 2)  Dividing derivatives:

       This form is widely used for derivatives of parametrically
       defined functions.  These are curves where a point (x,y)
       is determined by a 3rd quantity (parameter). So, x  and y
       are given in terms of this parameter.  Popular parameters
       are time t and angles q (theta). Values for the parameter will
       determined the location of the point (x,y) on our curve. In
       calculus II and III, you will study vector functions which will
       have parametric components. This gives a point much
       freedom of motion since the curve is not required to be a
       function of x. (the coordinates x & y are functions of the
       parameter). A popular application in calculus III would be
       the analysis of space curves.

       This form also allows us to solve interesting rate of change
        problems. (finding rate of change of one quantity with respect
        to another quantity).  Here is an example of both the parametric
        form and the rate of change application using form #2 of the
        chain rule.

 Example:       Given parametric equations  x = t² and y = sin(t),
                     find dy/dx.

                     Note that dx/dt = 2t and dy/dt = cos(t) are easily found.
                    To get dy/dx, we divide   (dy/dt) by (dx/dt).  We treat
                     the derivatives as if they were fractions with separate
                     numerators & denominators (they are not, not yet..we
                     will define dx and dy separately, coming soon...they will
                     be called differentials).

                     So, for this problem,  dy/dx = [cos(t)] / 2t

  Example:     Find the instantaneous rate of change of the volume of
                   an expanding cube with respect to (wrt) its surface area
                   at the instant when the edge is 3 inches.

                  The formulas here are  V = x³ and S = 6x², where x
                   represents the edge of the cube.  
                   
                   We want  dV/dS at x = 3"

                   We can easily get  dV/dx = 3x² and dS/dx = 12x,
                   so, dV/dS = 3x² / 12x = (1/4) x.
                   At x = 3, dV/dS = 3/4 inches cube per square inch

 ~Note:   Using this technique, we can find the instantaneous rate
              of change of any quantity wrt another quantity.  Just form
              two equations with the letters of your choice by calling
              each a single variable.

              Example:   find the rate of change of   x² + x -7  wrt  x /(x-1)
                               Let A = x² + x -7 and B = x /(x-1) then proceed
                               as in the last example. Your letters could be much
                               more creative. (compute dA/dB)