- The graph of y = f'(x) is given. Note: This is the derivative, not the function itself. Identify the points where the derivative is increasing and decreasing most rapidly, the points where the original function is increasing and decreasing most rapidly, where the original function has a horizontal tangent, and where there is a relative maximum or minimum in the original function.
- The graph of a function is given. Sketch the derivative of the function using the same coordinate system. Three parts. Check out the Derivative Plotter at Flash and Math.
- Find the second derivative of a function.
- Find the derivative using the limit definition. You will need to have the limit definition memorized.
- Use the table of values to find the derivatives at the indicated point. See below for an example. Four parts.
- Related rate story problem. Two parts.
- Differentiate. These come from throughout the chapter. You will need to know the basic functions, trigonometric functions, absolute values, piecewise functions; the product, quotient, and chain rules; implicit differentiation. Ten parts.
- Find the limits analytically and show work. These are similar to question 6 on the first exam. Two parts.

- Problems are generally a little harder than those in the textbook.
- Don't spend too long on any one problem. If you get stuck, make a note to come back and then move on.
- Concentrate on the problems that are worth the most points.

x | f(x) | g(x) | f'(x) | g'(x) |
---|---|---|---|---|

-3 | 2 | 5 | -1 | 4 |

-1 | 4 | -3 | 1 | 2 |

2 | -1 | 4 | -3 | -4 |

4 | -6 | -3 | 2 | 5 |

To make sure you understand how to read the table, f(-3) means find the -3 in the x column and then go over to the f(x) column to get f(-3) = 2.

- For interpretation of problems, (f-g)'(2) means find the derivative of f(x) and g(x) and evaluate the derivative at x = 2. (f-g)'(2) = f'(2) - g'(2) = -3 - (-4) = 1
- Find \( \left . \left ( g [f(x)] \right )' \right |_{x=-3} \). This is the chain rule, the derivative of the outside function g, evaluated at the inside function f(x), times the derivative of the inside function. \[ g'[f(-3)] \cdot f'(-3) = g'[ {\color {red}2} ] \cdot (-1) = (-4)(-1) = 4 \]
- Find \( \left ( g \, / f \right )'(-1) \). This is the quotient rule ... lo-de-hi minus hi-de-lo, square the bottom, there you go. Notice that the "low" (bottom) function is f and the "high" (top) function is g. \[ \frac {f(-1) g'(-1) - g(-1) f'(-1)}{( \,f(-1)\, )^2} = \frac {(4)(2) - (-3)(1)}{(4)^2} = \frac{11}{16}\]
- Find \( \left . \left ( 16x^2f^3 \right)' \right |_{x=4} \). The big picture here is the product rule, \( 16x^2 \) and \( f^3(x) \). Note that you could pull the 16 out as it is a constant. When finding the derivatives inside the product rule, you will need to use the power rule and then the chain rule. \[ \begin{align} \left . \left ( 16x^2f^3 \right)' \right |_{x=4} &= \left . (16x^2)(f^3)' + (16x^2)'(f^3) \right |_{x=4}\\ &= \left . 16x^2(3 [f(x)]^2 f'(x)) + 32x[f(x)]^3 \right |_{x=4}\\ &= \left . 48x^2 [f(x)]^2 f'(x) + 32x [f(x)]^3 \right |_{x=4} \\ &= 48({\color{red}4})^2 [f({\color{red}4})]^2 f'({\color{red}4}) + 32({\color{red}4}) [f({\color{red}4})]^3 \\ &= 768 (-6)^2(2) + 218(-6)^3 \\ &= 55296 - 47088 \\ &= 8208 \\ \end{align}\]

# | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | Total |
---|---|---|---|---|---|---|---|---|---|

Pts | 10 | 9 | 5 | 6 | 12 | 8 | 50 | 6 | 106 |