Section 6.1 Polynomial Derivatives, Product Rule, Quotient Rule.

Section 6.1Polynomial Derivatives, Product Rule, Quotient Rule

Limit Definition of the First Derivative

h 0

f x h f xGiven f x , f ' x lim

h

Symbols for the Derivative with Respect to x

dy df ' x f x y '

dx dx

Differentiation – the process of calculating derivatives

Polynomial RuleProduct RuleQuotient Rule

If dcxbxaxxf nnn ...21

THEN

0...21' 321 nnn xncxnbnaxxf

2f x 3x 4x 5

f ' x 6x 4 3 2f x 7x 4x 11

2f ' x 21x 8x

11 9 6 4f x x x x x

10 8 5 3f ' x 11x 9x 6x 4x

50 25 12 10f x x x x x

49 24 11 9f ' x 50x 25x 12x 10x

Now try these…..

4 2

6 3

4 1/ 2

4 4 2

f x 6x 3x x 5

f x 7x x 2x 1

f x 11x 3x 4x 1

f x 7x 5x 3x 2x 1

3

5 2

3 3 / 232

5 3 3

f ' x 24x 6x 1

f ' x 42x 3x 2

f ' x 44x x 4

f ' x 28x 20x 6x 2

Find the equation of the line tangent to 4 3 2y x 3x x x 1at x 2

3 24At x , y 2 3 2 2 12 02 4

3 24x 9x 2xdy

dx1

2

3 2

x 4 2 9 2dy

71dx

2 2 1

1 1my xy x

7y 40 2x1

If f x F S where

•F represents a function (first factor)•S represents a function (second factor)

F'S 'f ' x S F

2 1f x 3x 6 2x

4

21f ' x 6x 2x 2 3x 6

4

2 5f x 2 x 3x 7 x

5 4 2f ' x 1 6x 7 x 5x 2 x 3x

32

1 1f x 3x 27

x x

1 2 3x x 3x 27

3 22 3 2

1 2 1 1f ' x 3x 27 9x

xx x x

NIf f x

D

then

N'DIf f ' x

2D D

D'N

3xf x

2x 1

2

3 2x 1 2 3xf ' x

2x 1

2x 1

f x3x 2

2

2

2x 3x 2 3 x 1f ' x

3x 2

3 2

7

3x 6x 1f x

2x 4x 2

2 7 6 3 2

27

9x 12x 2x 4x 2 14x 4 3x 6x 1f ' x

2x 4x 2

2y x 4

y ' 2x

y '' 2

14 12 6y x x 2x

13 11 5dy14x 12x 12x

dx

212 10 4

2

d y172x 132x 60x

dx

2 1f x 3x 6 2x

4

f x FS

f ' x F'S S'F

21f ' x 6x 2x 3x 6 2

4

32

1 1y 3x 27

x x

1 2 3y x x 3x 27

32 1 223y ' 3x 27 9x1x 2x x x

32

23 2

1 2 1 1y' 3x 27 9x

xx x x

3x

1f x

2x

2

2x 1 2

2x 1

3f x

3x'

2

6x 3 6x

2x 1f ' x

2

3

2 1f x

x'

f x x

1/ 2f x x

1/ 21f ' x x

2

1/ 2

1f ' x

2x

1f ' x

2 x

3 2f x x

2/ 3f x x

1/ 32f ' x x

3

1/ 3

2f ' x

3x

3

2f ' x

3 x

f x 2 x

1/ 2f x 2x

1/ 21f ' x 2 x

2

1/ 2

1f ' x

x

1f ' x

x

POSITION s(t)

VELOCITY v(t)

ACCELERATION a(t)

DIFFERENTIATE

INTEGRATE

Given 2s t 3t 4t 7 :

a) Find v(4)

24 0v 6 4 4

b) Find a(2)

v t 6t 4

a t 6

a 2 66

v t 6t 4

Given 3 2s t 3t t 4t 7

a) Find the initial velocity of the object.

Initial velocity is the velocity at t = 0.

2v t 9t 2t 4

2v 0 9 0 2 0 4 4

b) Find the value of the acceleration at t = 2.

2v t 9t 2t 4

a t 18t 2

a 2 18 2 342