Warmup 10/17/16 Objective Tonight’s Homework Has anything happened in your life that has given you cause to doubt God’s existence? Objective Tonight’s Homework To learn how we take the derivative of two multiplied functions Prove the product rule for any functions. pp 172: 3

Homework Help Let’s spend the first 10 minutes of class going over any problems with which you need help. 2

Notes on the Product Rule If we have two functions multiplied, how do we take the derivative?

Notes on the Product Rule If we have two functions multiplied, how do we take the derivative? Example: find dy/dx of y = 3x•x2 4

Notes on the Product Rule If we have two functions multiplied, how do we take the derivative? Example: find dy/dx of y = 3x•x2 Let’s do this the long way. lim 3(x+h)(x+h)2 – 3x•x2 h h  0 5

Notes on the Product Rule If we have two functions multiplied, how do we take the derivative? Example: find dy/dx of y = 3x•x2 Let’s do this the long way. lim (3x + 3h)(x2 + 2xh + h2) – 3x3 h 3(x+h)(x+h)2 – 3x•x2 h h  0 6

Notes on the Product Rule If we have two functions multiplied, how do we take the derivative? Example: find dy/dx of y = 3x•x2 Let’s do this the long way. lim (3x + 3h)(x2 + 2xh + h2) – 3x3 h 3x3 + 6x2h + 3xh2 + 3hx2 + 6xh2 + 3h3 – 3x3 3(x+h)(x+h)2 – 3x•x2 h h  0 7

Notes on the Product Rule 3x3 + 6x2h + 3xh2 + 3hx2 + 6xh2 + 3h3 – 3x3 h h(6x2 + 3xh + 3x2 + 6xh + 3h2) 8

Notes on the Product Rule 3x3 + 6x2h + 3xh2 + 3hx2 + 6xh2 + 3h3 – 3x3 h h(6x2 + 3xh + 3x2 + 6xh + 3h2) lim (6x2 + 3xh + 3x2 + 6xh + 3h2) h  0 9

Notes on the Product Rule 3x3 + 6x2h + 3xh2 + 3hx2 + 6xh2 + 3h3 – 3x3 h h(6x2 + 3xh + 3x2 + 6xh + 3h2) lim (6x2 + 3xh + 3x2 + 6xh + 3h2) 6x2 + 3x2 h  0 10

Notes on the Product Rule 3x3 + 6x2h + 3xh2 + 3hx2 + 6xh2 + 3h3 – 3x3 h h(6x2 + 3xh + 3x2 + 6xh + 3h2) lim (6x2 + 3xh + 3x2 + 6xh + 3h2) 6x2 + 3x2 This is all well and good, but we need a pattern. Compare this answer to our original function. original: y = 3x•x2 derivative: 6x2 + 3x2 h  0 11

Notes on the Product Rule original: y = 3x•x2 derivative: 6x2 + 3x2 Our pattern is this: If “u” and “v” are functions, the derivative of u•v is: Product Rule (uv)’ = uv’ + vu’ 12

Notes on the Product Rule original: y = 3x•x2 derivative: 6x2 + 3x2 Our pattern is this: If “u” and “v” are functions, the derivative of u•v is: Product Rule (uv)’ = uv’ + vu’ An easy way to remember this is with the saying: “First dee second plus second dee first” 13

Notes on the Product Rule Example: find the derivative of.. x3 • sin(x) x3 is “U” and sin(x) is “V”. So according to our new rule, the derivative will be… UV’ + VU’ (x3)(cos(x)) + (sin(x))(3x2) x3cos(x) + 3x2sin(x) 14

Group Practice Look at the example problems on pages 169 through 172. Make sure the examples make sense. Work through them with a friend. Then look at the homework tonight and see if there are any problems you think will be hard. Now is the time to ask a friend or the teacher for help! Prove the product rule for any functions & pp 172: 3 15

Exit Question dy/dx of y = x2•ln(x) = ? a) x2•1/x + ln(x)•2x b) x2•ln(x) + 2x•1/x c) x2•1/x • ln(x)•2x d) x2•ln(x) • 2x•1/x e) x2 + 1/x • ln(x) + 2x f) x2 + ln(x) • 2x + 1/x