MATH 1351 TI-85 EXERCISE VII

Derivatives as the limit of the difference quotient

In Section 1 of Chapter 3 we learned that the derivative of a function f at the input value x is the limiting value of the slopes of the chords drawn from ( x , f(x) ) to nearby points on the graph. This slope from ( x , f(x) ) to the nearby point ( x + h , f( x + h) ) is the difference quotient

DQ:	( f(x + h) - f(x) ) / h

for small positive and negative values of h. And its' limiting value we defined to be the slope of the tangent line to the graph at ( x , f(x) ).

Also in Section 1, we saw how to find a formula for this derivative in terms of the value x for some very simple functions. Here we will try to conjecture what such a derivative formula might be for a much more sophisticated function; namely, we'll consider the function f(x) = ln (x).

We'll investigate the limit of the difference quotient at the three values x = 1, 2, & 3.

1. Compute each of the difference quotients for f(x) = ln (x) directly:

for x = 1 & h = 10^-10	DQ =

for x = 1 & h = -10^-10	DQ =

for x = 2 & h = 10^-10	DQ =

for x = 2 & h = -10^-10	DQ =

for x = 3 & h = 10^-10	DQ =

for x = 3 & h = -10^-10	DQ =

2. Sketch the graph of each DQ for each value of x = 1, 2, & 3 and use ZOOM and TRACE to estimate the value near 0. (The variable in each DQ is h, but the TI-85 recognizes only x as an independent variable. So we have to replace h in the above DQ with x. This shouldn't be too confusing since the x in the above DQ is to be replaced with the numbers 1, 2, & 3 anyway.)

So the graph of DQ corresponding to the point ( 1 , ln(1) ) is the graph of

y = ( ln ( 1 + x) - ln (1)) / x	and	limit as x -> 0 ~

At the point ( 2 , ln(2) ) the graph is that of

y = ( ln ( 2 + x ) - ln (2)) / x	limit as x -> 0 ~

At the point ( 3 , ln(3) ) the graph is that of

y = ( ln ( 3 + x ) - ln (3)) / x	limit as x -> 0 ~

3. Finally, simply sketch the graph of y = ln(x) in the ZDECM window. Choose MATH from the screen menu, then choose dy/dx from the MATH screen menu. TRACE is automatically activated. Move the cursor to x =1 and press ENTER. Do the same for each of x = 2 and x = 3. We obtain dy/dx = at x = 1, dy/dx = at x = 2, and dy/dx = at x = 3.

What appears to be the relationship between the value of x and that of dy/dx for y = ln (x)?