1. The Product Rule

2. Ideally, we can operate on expressions in factored form. Expanding expressions just to work with them seems time consuming. f(x) = (3x 2 – 1)(x 3 + 8)

3. Given h(x) = f(x)g(x), where f(x) = (x 2 + 2) g(x) = (x + 5) Does h’(x) = f’(x)g’(x) (it would be nice….)

4. h(x) = (x 2 +2)(x + 5) h(x) = x 3 + 5x 2 + 2x + 10 differentiate h’(x) = 3x 2 + 10x + 2 or f(x) = (x 2 + 2) f’(x) = 2x g(x) = (x + 5) So h’(x) = (2x)(1) = 2x ? (nope…) g’(x) = 1

5. The Product Rule If h(x) = f(x)g(x), then h’(x) = f(x)g’(x) + g(x)f’(x)

6. Differentiate (again…) h(x) = (x 2 + 2)(x + 5) h’(x) = (x 2 + 2) + (x + 5) = (x 2 + 2)(1) + (x + 5)(2x) = x 2 + 2 + 2x 2 + 10x = 3x 2 + 10x + 2

7. Homework

8. The Power of a Function Rule for Positive Integers If f(x) = [g(x)] n, then f’(x) = n[g(x)] n-1 g’(x)

9. Differentiate h(x) = (x 2 + 3x + 5) 6

10. homework Pg 145 [1,2,4] odd 5, 6a, 7a (use example 4)