Scribe Post 9/28

Today we began Unit 2: Derivatives
We spent most of class classifying the four ways we will be looking at the derivative:

- Verbally
- Numerically

- Graphically
- Algebraically

Verbal
 When we talk about the derivative, we will refer usually to the instantaneous rate of change, or the instantaneous rate. These both refer to the slope of the tangent line.

Numeric
 Numeric derivatives aren't a particularly sexy way of finding the instantaneous rate. They involve looking at a table of values around a point and finding the difference quotient (or ) for a very small . For example, given this data table:

Fig. a

We can find what is very close to the instantaneous rate at x = 3 by finding .

 A better way to find the numeric derivative is to use a symmetric difference quotient, where, for some function for which we want to find , instead of finding the difference quotient between and , where h is some very small value, we find the difference quotient between and .
Your calculator can do this for you using the nDeriv() function. nDeriv() has four parameters: the function you wish to differentiate, the variable that you are differentiating with respect to (for now this will always be x), the value, , at which you wish to find the derivative, and some small number h, which will be the interval above and below that the calculator will find the difference quotient between. To define nDeriv more explicitly:

If you only input three arguments, the calculator will use 0.001 for h.

While numeric differentiation is very easy to do with technology, it lacks elegance, and most of the time, there will be a much cooler way to do it.

Graphic
Graphic differentiation involves actually finding the tangent line, then finding the slope of that line. We used Geogebra to graph the following function:

Next we added a fixed point, A, at (2, 4) on the parabola, a moveable point, B, also on the parabola, and a line between A and B. This line is called the secant line.

Next, we dragged the point B closer to point A and noticed that as B got pretty close to A, the secant line began to resemble a tangent line.

However, when we set points A and B equal, the line disappeared, because the slope of the line would be .

Next, we looked at a much cooler function:

 

We used one of Geogebra's built in functions, Tangent[], to draw a line tangent to g at a moveable point A (Tangent[] has two parameters: a point, and a function).

Next, we used Geogebra's Slope[] function on our tangent line, and created a point, B, (x(A), slope[<tangent line>]).

 
Finally, we turned on trace for point B and animated A.

By guess and check, we found that the locus of points described by tracing B could be modeled perfectly with the function

 
Weird, that the natural log should be related to a hyperbola...

EDIT: Having just learned about the derivatives of exponential and logarithmic functions, I thought it would be appropriate to show that the derivative shown graphically above is actually the derivative:

 This is just a normal chain rule problem, but we could also, if we forgot the derivative of the natural log, find the same answer by differentiating implicitly:

 And finally by substituting back in y from the original equation:

 Q.E.D.

Algebraic
At this point, it became clear that our current definition for the tangent was insufficient for finding the derivative algebraically. However, during our graphical investigation, while looking at the function , we saw that a secant line defined by two points, A and B, both on f. As B approached A, the line looked more and more like a tangent line. Therefore, the slope of the tangent line, which is what we are really after, is the difference quotient of A and B as B goes toward A. The explicit definition is, for and :

To illustrate this further, we examined problem 16 from the test, which told us to find:

Mr. O'Brien pointed out that this was just the derivative of   at 2.

 

Because when you take the derivative, it will always result in an indeterminate limit, we have to screw around with the expression before we can find its value by substitution:

So the derivative at 2 of is , but more interestingly, on line six above, just before we substituted in the 2 for x, we came up with the equation:

We don't have to substitute any value for c into this equation. It is no longer indeterminate, so we can simply substitute in c for x:

And from there, what's to stop us from renaming c as x?

Now, instead of having to go through a long series of calculations simply to find the slope of g at one point, we have a handy function that describes it. Cool, right? But didn't Mrs. Damien show us how to do this to any function in the entire world with just a few simple steps? What's the point of this? Unfortunately, the "Damien Rule" doesn't always work: cannot be differentiated using it; nor can any of the trig functions. Clearly, the Damien Rule is very useful, but not comprehensive. We worked through the below problem to show the Damien Rule in action:

Damien Method

They yield the same result, but between steps eight and nine on the first method, we had to do that weird substitution of x for c again. This is because the formula for the derivative that we have been using () is actually the formula for evaluating the derivative at a point. There is a second formula for finding the function of the derivative:

To show that this indeed is also the derivative, we found the derivative of h(x) one more time:

Much more better.

For an in depth explanation of this second definition of the derivative, go here and select the last clip. It wouldn't be an awful idea to just watch the full first lecture, though.

The next scribe will be...
CALVIN.