The Product Rule. Factoring has become an indispensible tool. Since we are always re-writing polynomials as a product, we should generate a rule for products…

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Presentation transcript:

The Product Rule

Factoring has become an indispensible tool. Since we are always re-writing polynomials as a product, we should generate a rule for products…

Also, 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)

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….) SO…

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

The Product Rule (used when the expression is in factored form) If h(x) = f(x)g(x), then h’(x) = f(x)g’(x) + g(x)f’(x)

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

Pg 93 [2,3,5,6] a,c,e 8,9,14,15,16

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)

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

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