3.3 Differentiation Rules Colorado National Monument Photo by Vickie Kelly, 2003 Created by Greg...
-
Upload
jada-mclain -
Category
Documents
-
view
213 -
download
0
Transcript of 3.3 Differentiation Rules Colorado National Monument Photo by Vickie Kelly, 2003 Created by Greg...
3.3 Differentiation Rules
Colorado National MonumentPhoto by Vickie Kelly, 2003
Created by Greg Kelly, Hanford High School, Richland, WashingtonRevised by Terry Luskin, Dover-Sherborn HS, Dover, Massachusetts
If the derivative of a function is its slope, then for a constant function, the derivative must be zero for all x.
0dc
dx
example: 3y
(3) or (3) 0d
ydx
The derivative of a constant function is zero.
We saw that if , .2y x 2dy
xdx
This is part of a larger pattern!
1n ndx nx
dx
examples:
f x x
01 1f x x
8y x
78y x
“power rule”
( )
!
d
dxd
dxd
dxwhere is some
u
u
u
u
functi n
c
f x
c
o
c
o
examples:
( )ncd
xdx
constant multiple rule:
5 5(7 ) 7 ( )d d
x xdx dx
47 5x 435x
( )ncdx
dx1ncnx
(Each term is treated separately)
sum and difference rules:
ud d d
dx d x
vv
x du ud d d
dx d x
vv
x du
4 12y x x
y
4 23 2 9y x x dy
dx
12 4x
!fu
whe
nct
re and
ions
reu v
of x
a
034x 312x
Example:Find the horizontal tangents of: 4 22 2y x x
34 4 0dy
x xdx
Horizontal tangents: (slope of y) = (derivative of y) = zero.
34 4 0x x
3 0x x
2 1 0x x
1 1 0x x x
0, 1, 1x
Substitute these x values into the original equation to generate coordinate pairs: (0, 2), (-1, 1) and (1,1)
2 0( 0)
1 0( 1)
1 0( 1)
y x
y x
y x
(So we expect to see two horizontal tangents, intersecting the curve at three points.)
… and write tangent lines:
4 22 2y x x
4 22 2y x x
2y
4 22 2y x x
2y
1y
4 22 2y x x
First derivative y’ (slope) is zero at: 1, 0, 1x
34 4dy
x xdx
…where y has slopes of zeroThe derivative of y is zero
at x= -1, 0, 1…
product rule:
d du dvu v v u
dx dx dx
Notice: the derivative of a product is not just the product of the two derivatives. (Our authors write this rule in the other order!)
This rule can be “spoken” as:
2 33 2 5d
x x xdx
d(u∙v) = v ∙ du + u ∙ dv
= (2x3 + 5x) (2x) + (x2 + 3) (6x2 + 5)
= (4x4 + 10x2) + (6x4 + 23x2 + 15)
= 10x4 + 33x2 + 15
u = x2 + 3v = 2x3 + 5x
du/dx = dv/dx =
2x + 02∙3x2 + 5
quotient rule:
2
du dvv ud u dx dx
dx v v
or 2
u v du u dvdv v
3
2
2 5
3
d x x
dx x
2 3x
u = 2x3 + 5x
v = x2 + 3
du =
dv =
22 3 5x
2 0x 2 23 6 5x x
2 2 33 6 5 2 5x x x x
2 2 33 6 5 2 5 2x x x x x
2 2 3
22
3 6 5 2 5 2
3
x x x x x
x
Higher Order Derivatives:
dyy
dx is the first derivative of y with respect to x.
2
2
y dy
d
d d d y
dx dx dxxy
is the second derivative
(“y double prime”), thederivative of the first derivative.
3
3
d d y
dy
x
y
x d
is the third derivative
We will see later what these higher-order derivatives might mean in “the real world!”
(“y triple prime”), the derivative of the second derivative.