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
[2^{nd}][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.)

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