difference_and_derivatives3

Differences and Derivatives



I. Symbols of Functions

y

“Rise over run” is a way to remember slope that avoids symbols. is a symbolic

x

model for slope that uses the idea of change but (still) avoids the use of functional

symbols. Calculus typically forces the issue and eventually obliges the learner to grapple

with functional notation. Today is that day.









C









B



y  y1x 

A





x  x x x  x









Before setting up the ratios, let’s first study the symbols for the coordinates.



Point A: horizontal coordinate = x  x ; vertical component = y1x  x 

Point B: horizontal coordinate = x ; vertical component = y1x 

Point C: horizontal coordinate = x  x ; vertical component = y1x  x 



Take a couple of minutes to graph and write the coordinates for points that are a distance

of 2x away from the center x .1





1

The points have horizontal coordinates x  2x and x  2x . The corresponding

vertical coordinates are y1x  2x and y1x  2x .

Now practice writing the “rise over run” for a few segments that connect any of these

points. One is done for you as a check



Points Rise y

Slope = 

Run x



 x, y1x  and  x  x, y1x  x  y1x  x   y1x 

x









Say each time you write a completed fraction for rise over



run. This is real algebra.









II. Creating Differences Data: Our 1st Rate of Change Graph



The “Difference Quotient tech handout” from Session 9 shows how to set up a single

ratio of differences on your home screen:



y15   y13 y15   y13

or

53 2



The difference is also the slope between two points on the graph of y  y1x  . To get

accurate estimates best we take points close together, and better yet if we let your

calculator do all of the work. In the steps below we somewhat arbitrarily choose

x  0.01 .

STEPS

1. Enter the function y1  2 x

2. Go to the lists -- LIST (on TI-73) and STAT/EDIT(TI-83) -- Clear out lists L1

and L2.

3. In L1 enter input values:

2, 1.75, 1.5, 1.25, 1, .75, .5, .25, 0,.25,.5,.75,1,1.25,1.5,1.75, 2

4. Arrow to the very top of L2 and enter the command:



L2=”(y1(L1 + .01)-y1(L1))/.01)”2





“L2 =” should appear in the bottom left hand corner of the screen, and you must

paste the name L1 as a symbol. Don’t type L and then 1. As a check on your

work, the first entry in L2 should be 0.173389.



5. Now create a scatter plot to graph L1 against L2 . Be sure that y1 is also active

so that both y1  2 x and its rate of change are plotted together.



Compare the graph of y1  2 x and the scatter plot of the slopes. How are they similar?

How are they different? What do you think the “formula” of slopes (scatter plot) might

look like? Hint: recall “Ah Has” from last week and the effect of coefficients.









III. More Differences Data: Another Rate of Change Graph

If you alter the base of your function y1, and repeat the process of part II, you would

again find that the rate of change graph is an exponential with a different y-intercept.

Since the intercept determines the appropriate coefficient to use in a formula, we can

make a whole set of formulas. Complete the following table of exponential derivative

formulas. Now this is REALLY









2

The quotation marks will make it so we don’t have to retype the formula when we

change y1 later in the handout. However, they can get us into trouble – since L2 is

defined in terms of y1, if there is no y1, the calculator will give an error

(ERR:INVALID). To find the quotation marks on the TI-73 push 2nd TEXT.

Exponential base Rate of change formula

2







3







4







5









2.7183









Can you make any conjectures about the derivatives of exponential functions?









3

This pattern requires a lot of expertise – even Fermat missed it. It had to wait until

Euler had done his work. Yes, you may recognize that 2.718 . . . is an approximation for





e !