Post on 16-Jan-2016
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