1. AP Calculus AB/BC 3.2 Differentiability, p. 109 Day 1

2. To be differentiable, a function must be continuous and smooth. Derivatives will fail to exist at: Corner – one-sided derivatives differ. Cusp – slopes of the secant lines approach −∞ from one side and ∞ from the other side.

3. Vertical tangent – the slopes of the secant lines approach −∞ or ∞ from both sides. Discontinuity – One of the one-sided derivatives is nonexistent.

4. Most of the functions we study in calculus will be differentiable.

5. Example 1:The function below fails to be differentiable at x = 0. Tell whether the problem is a corner, a cusp, a vertical tangent, or a discontinuity. Since we want to differentiate at x = 0, we know at x = 0, y = 1 by the definition. But, tan −1 0 = 0. So, therefore the problem is discontinuity because there are two values of y at x = 0.

6. Derivatives on a Calculator, p. 111 For small values of h, the difference quotient often gives a good approximation of f ′(x). The graphing calculator uses nDERIV to calculate the numerical derivative of f. A better approximation is the symmetric difference quotient which the graphing calculator uses.

7. Example 2: Computing a Numerical Derivative Compute the numerical derivative of f(x) = 4x – x 2 at x = 0. Use h = 0.001. = 4 4(x + h) (x + h) 2 4(x − h)(x − h)2(x − h)2

8. Derivatives on the TI-83/TI-84/TI-89: You must be able to calculate derivatives with the calculator and without. So, you need to do them by hand when called for. Remember that half the test is no calculator.

9. Example 3 : Find at x = 2. 8: nDeriv ( x ^ 3, x, 2, 0.001 ) ENTER returns MATH Note: nDeriv ( x ^ 3, x, 2, 0.001 ) is the same as nDeriv ( x ^ 3, x, 2).

10. Warning: The calculator may return an incorrect value if you evaluate a derivative at a point where the function is not differentiable. Examples: returns nDeriv ( 1/x, x, 0 ) nDeriv ( abs(x), x, 0 )

11. Graphing Derivatives Graph: What does the graph look like? This looks like: Use your calculator to evaluate: −10 ≤ x ≤ 10 −10 ≤ y ≤ 10 

12. AP Calculus AB/BC 3.2 Differentiability, p. 109 Day 2

13. There are two theorems on page 113: Theorem 1: Differentiability Implies Continuity If f has a derivative at x = a, then f is continuous at x = a. Since a function must be continuous to have a derivative, if it has a derivative then it is continuous.

14. Between a and b, must take on every value between and. Theorem 2: Intermediate Value Theorem for Derivatives If a and b are any two points in an interval on which f is differentiable, then f ′ takes on every value between f ′(a) and f ′(b). Example 1

15. Example 2:Find all values of x for which the function is differentiable. For this particular rational function, factor the denominator. The zeros are at x = 5 and x = −1 which is where the function is undefined. Since f(x) is a rational function, it is differentiable for all values of x in its domain. Therefore, f(x) is not differentiable at x = 5 or x = −1 since 5 and −1 are not in the domain of f(x). 