Calculator: TI-83/84.
The numerical derivative estimates the rate of change of one variable relative to another in a "short" interval.
Let y = f(x), where f is an expression specifying how y depends on x, and let c be a value of x. In order to estimate the rate of change of y relative to x around the value c (x = c), we choose a small positive number h, and we compute:
(f(c+h) - f(c-h)) / (2*h).
The TI-83/84 (as well as other graphing calculators) provides a mathematical function that computes the numerical derivative, nDeriv.
nDeriv(f,x,c,h) = (f(c+h) - f(c-h)) / (2*h)
If h is not specified, the value, h = 0.001 is used, so
nDeriv(f,x,c) = (f(c+0.001) - f(c-0.001)) / 0.002
Task 1.
Compute the numerical derivative of y = (x - 1)3 + x, relative to x, for around the value x = 1, using h = 1, 0.1, 0.01, 0.001, and 0.0001.
Solution.
Enter,
\Y1=(X-1)3+X
\Y2=nDeriv(Y1,X,1,H)
From the home screen run:
h [ENTER] for h = 1, ... , .0001.
Ans → H:Y2 [ENTER] Use [2nd][ENTRY] to retrieve this instruction.
Table of answers.
| H = | nDeriv = |
| 1 | 2 |
| 0.1 | 1.01 |
| 0.01 | 1.0001 |
| 0.001 | 1.000001 |
| 0.0001 | 1.00000001 |
Task 2.
Edit Y2,
...
\Y2 = nDeriv(Y1,X,X) Now h = 0.001, and the value x = c is a variable.
Set the window, Xmin=-1, Xmax=3, Ymin=-10, Ymax=15, and [GRAPH].
For which value of x is the rate of change of y relative to x the smallest?
(Use [TRACE] to find the answer.)