~NEWTON'S DIFFERENCE QUOTIENT~

cockatoo


~the definition of f '(x) = lim  [f(x+h) - f(x)] / h
                                        h→0

~note:  Dx is also used in place of h.

~This definition is referred to as the 4 step method or process in many textbooks.

~note: this definition gives the derivative as a function of x.
          So, it describes the slopes of the tangent lines for all x
           in the domain of f.  If you are interested in a particular
           point on the function, you must substitute the x-value
           (sometimes the y value also) into the derivative.

~Example:  given f(x)=x2,  compute f '(x)

          f '(x) = lim  [(x+h)2 - x2] / h
                       h→0

                = lim  [x2+2hx+h2 - x2] / h
                     h→0

                   = lim  (2hx+h2)/h = lim  (2x+h)=2x
                      h→0                  h→0

      So,  f '(x) = 2x     (substituting the x value of a point,
                                    will give the slope of the tangent
                                    line at that point)

~the definition of f '(a):  use this to examine the derivative
                                        at a specific point where x=a.

              f '(a) =  lim  [f(x) - f(a)] /(x-a)
                          x→a

   let's examine f '(3) for the above function.

             f '(3) =  lim  [x2 - 32] / (x-3)
                          x→3

                      =  lim  [(x-3)(x+3)] / (x-3)
                         x→3

                      =  lim  (x+3) = 6
                         x→3

~note: we could have computed f '(x) using the first
           definition then substituted x=3 into that expression,
           however, this definition is very easy to use, if just
           interested in the derivative at one place. Many
           questions deal with whether or not a given function
           is differentiable at a given point & this definition is
           is best to use for that type of a problem.

Note:  See  Derivative by definition  for another example



WHAT DOES A DERIVATIVE MEAN?


~First of all, you'll need a function to work with. This will serve as our model. It will take the form of an equation relating the dependent & independent variables. This could be given in functional notation or in dependent variable notation.

~For example, if G represents your grade on exam #1 and t represents your study time in hours, then G(t) would be that function. Here we would be looking at G  vs. t, graphically.

~So, G=f(t). We could use dependent variable notation with G or functional notation with f(t).

~Also, G(t) could be used for f(t).

~The derivative of G with respect to t can be denoted several ways: 

G '(t), f '(t), dG/dt, G ', f ' would be the popular ways.

~The derivative gives us how the dependent variable G (grade on exam 1) is changing at the INSTANT when the independent variable t (study time) is at a given value.

~The units for the derivative would be dependent variable units/each unit of the independent variable.

 ~For Grade vs. Study time, the derivative would have units like  2.5 pts/hr , or -3.7 pts/hr, or 0 pts/hr all at a given value of t. In the first case, your grade is increasing by 2.5 pts/hr at a specific value of t. In the 2nd case, your grade on exam 1 would be decreasing by 3.7 pts/hr at some value of t.  In the 3rd case, your grade would be neither increasing nor decreasing at some instant of time. (0 pts/hr).

~Don’t confuse G'(5) with G(5). The first is the derivative at t=5 while the 2nd is the functional value, G at t=5. The functional value is your grade at t=5 hours, while the derivative is how it’s changing at t=5 with respect to the amount of study time.

~Graphically, G '(t) would represent the value of the slope of a tangent line drawn to the curve, G=f(t), at that specific value of t while G(t) would be the functional value or level of the curve at that value. If you know G(t) for a given t, you have a point on the graph. For example, G(5)=78 gives the point (5,78) on the graph of G.

~Let’s say you have a model that would estimate your grade with respect to time studied as follows:  G(t)=10t-.25t2.  

~Using the definition of G'(t) (4 step method) or a derivative formula, we would get,  G'(t)=10-.50t.

At t=5 hours (time spent studying), G(5)=50-.25(25)= 50-6.25 = 43.75 (discouraging)
After 6 hours, G(6)=60-.25(36)=60-9=51 (still discouraging, but improving)
After 7 hours, G(7)=70-.25(49)=70-12.25=57.75 (improving)
After 10 hours, G(10)=100-.25(100)=100-25=75 (solid C)
After15 hours, G(15)=150-.25(225)=150-56.25=93.75 (an A) (nice)
After 28 hours, G(28)=280-.25(784)=280-196=84 (B)
(too much study time & was too tired at exam time)

~Now, let’s look at the derivatives at these times.
G '(5)=7.5 pts/hr (your grade would be increasing at this rate at t=5)
G '(6)=7 pts/hr (increasing G also at t=6, but not as much)
G '(7)=6.5 pts/hr (still increasing at t=7, but a little slower)
G '(10)=5 pts/hr (slower increase at t=10)
G '(15)=2.5 (continuing a slower increase at t=15 for G)
G '(28)= - 4 pts/hr (your grade will be decreasing at t=28)
(this could be due to certain factors like lack of sleep thus having a lack of concentration during test time)

~The graph of G is parabolic (opening down). It reaches its maximum at t=20. For this exact value of t, your test grade would be 100. This would be your optimum study time. However, remember that this is a mathematical model & models rarely match up with reality.