## November 15, 2006

### The Product Rule

Today we found the Product Rule for derivatives and as usual we had white board questions!

1. f(x) = x3 + 2x2 - 8x + 1
Find the equation of the tangent line at x=2

f(2) = 1

f'(x) = 3x2 + 4x - 8
f'(2) = 3(2)2 + 4(2) - 8
= 3(4) + 8 - 8
= 12 + 8 - 8
= 12

equation: y - 1 = 12(x-2)

2. g(x) = x3/2 - x1/2
Find the equation of the tangent line x=4

g(4) = 6

g'(x) = (3/2)x - (1/2)x
g'(4) = (3/2)x1/2 - (1/2)-1\2
= (3/2)(4)1/2 - (1/2)-1\2
= 11/4

equation: y - 6 = 11/4(x-4)

3. Multiply
(x + 3)(2x +5)

I know that everyone can easily do this problem but this is going to help us derive the product derivative rule. I wasn't taught how to multiply by using "the area model of multiplication" and it probably wouldn't have changed how i multiplied but it will serve its purpose in the next question. Mr.K showed us an example with (x + 3)(2x +5)
= 2x2 + 11x + 15
With this example, it helped me understand how to find the derivative rule of F(x)=f(x) g(x).

The first thing we had to do was find the areas of I, II, and III

Area I: [g(x+h) - g(x)]f(x)

Area II: [f(x+h)-f(x)]g(x)

Area III: [f(x+h)-f(x)][g(x+h)-g(x)]

With these areas, we were able to derive the product rule of the derivative.

The Product Rule of the Derivative is f(x)g'(x) + g(x)f'(x).

No Homework today. My work here is done. Jann, I choose you!