- For the accompanying position versus time graph, find the average velocity over an interval, the places where the instantaneous velocity is zero, the times at which the instantaneous velocity is a maximum or minimum, and the instantaneous velocity at a specific time. Look at problem 3.1.2
- Sketch the graph of the derivative of the function whose graph is shown. Look at problems 3.2.25-26.
- Match the graphs of the functions with the graphs of their derivatives. Eight parts. Look at problem 3.2.23
- Find the derivative of algebraic functions that do not require the use of the chain rule. You need to know the power, product, and quotient rules. Four parts. You may wish to multiply or divide before differentiating. Look at problems 3.3.1-20.
- Find f'(x) for trigonometric functions. Three parts. Look at problems 3.4.1-18.
- Find f'(x) using the chain rule. Three parts. Look at problems 3.5.7-40.
- The values of f, f ', g, and g' are given in a table. Use the rules of derivatives to find the derivatives as the specified point. Seven parts. Look at problems 3.3.39-42; 3.4.1-2, 5-6, 69; 3.summary.10 or see example below
- Find the differential dy. Look at problems 3.8.37-40.
- Use a local linear approximation to estimate the value given. Find the relative error in the estimation. Look at problems 3.8.19-27.
- Find a higher order derivative. Look at problems 3.3.45-48.
- Use implicit differentiation to find dy/dx. Look at problems 3.6.11-20.
- Use the limit definition of the derivative to find the derivative of the given function. Also note that this is a problem where I will be grading work more than the answer. Be sure you write the limit in every time until you actually take the limit. Look at problems 3.2.15-20 but use f'(x) = limit as h->0 of [ f(x+h)-f(x) ] / h.

- The problems are similar to those from the textbook.

x | -1 | 0 | 1 |
---|---|---|---|

f(x) | 1 | -1 | 3 |

g(x) | 2 | 1 | -1 |

f'(x) | 5 | 2 | 1 |

g'(x) | -2 | 3 | 0 |

- Find the derivative of [ f(x) g(x) ] |
_{x=1}. Since (fg)' = fg' + f'g, this becomes f(1)g'(1) + f'(1)g(1). Looking in the table to find those values, we get (3)(0) + (1)(-1) = 0 - 1 = -1. - Find the derivative of [ f
^{2}(x) / g(x) ] |_{x=0}. The big picture to this problem is a quotient, so we'll use the quotient rule. Within the derivative of the numerator part, we'll have to use the chain rule to find the derivative of f^{2}(x) . Since f^{2}(x) is*something*squared, the derivative is*2 * something * the derivative of something*. That is, it becomes 2 * f(x) * f'(x). The quotient rule is the denominator, g(x), times the derivative of the numerator, 2 f(x) f'(x), minus the numerator, f^{2}(x), times the derivative of the denominator, g'(x), all divided by the denominator squared, [g(x)]^{2}. This becomes [ g(x) * 2 f(x) f'(x) - f(x) * g'(x) ] / [ g(x) ]^{2}. Evaluating this at x=0, we get [ 2 g(0) f(0) f'(0) - f(0) g'(0) ] / [ g(0) ]^{2}= [ 2 (1) (-1) (2) - (-1) (3) ] / (3)^{2}= ( -4 + 3 ) / 9 = -1/9

# | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | Total |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|

Pts | 8 | 5 | 8 | 16 | 12 | 12 | 14 | 4 | 6 | 5 | 5 | 5 | 100 |

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Last updated
March 4, 2003 11:56 AM