the derivative-instantaneous rate of change the derivative of a function, f at a specific value of...
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The Derivative-Instantaneous rate of change
0( ) ( )
( ) limhdy f a h f a
f adx h
0( ) ( )
( ) limhdy f x h f x
f xdx h
The derivative of a function, f at a specific value of x, say a is a value given by:
The derivative of a function, f as a function of x, is called f (x) and is given by:
0( ) ( )
( ) limhdy f x h f x
f xdx h
2( ) 3 4 7f x x x
2
2 2
( ) 3( ) 4( ) 7
3 6 3 4 4 7
f x h x h x h
x xh h x h
Find the derivative of
2 2 2
0(3 6 3 4 4 7) (3 4 7)
limhx xh h x h x x
h
2
0 06 3 4 (6 3 4)
lim limh hxh h h h x h
h h
0lim 6 3 4 6 4h x h x
( ) 6 4f x x
Related problems 2( ) 3 4 7f x x x
1) Find the slope of f (x) at x = 3, x = -2
3) Find the point on f (x) for which the slope is 2
4) Find the point for which f (x) has a horizontal tangent line
2) Write the equation of the tangent line at x = -2
Solutions 2( ) 3 4 7f x x x
1) Find the slope of f (x) at x = 3, x = -2
2) Write the equation of the tangent line at x = -2
( ) 6 4f x x so
( 2) 6( 2) 4 16f
(3) 6(3) 4 14f
1 1( )y y m x x 2( 2) 3( 2) 4( 2) 7 27f
27 16( 2)y x
27 16( 2)y x 16 5y x
use the point-slope formula
Find the value of y
Solutions 2( ) 3 4 7f x x x
3) Find the point on f (x) for which the slope is 2
4) Find the point for which f (x) has a horizontal tangent line
( ) 6 4f x x 6 4 2
1
x
x
2
3x
2(1) 3(1) 4(1) 7 6f The point is (1, 6)
6 4 0x
22 2 2 17( ) 3( ) 4( ) 73 3 3 3
f The point is (2/3, 17/3)
Figure 2.7: Derivatives at endpoints are one-sided limits.
Derivatives at Endpoints are one-sided limits.
How a derivative can fail to exist
A cornerA vertical tangent
A discontinuity
Which of the three examples are the functionscontinuous?
The graph of a function
The graph of the derivative (slope) of the function
Where f(x) is increasing(slope is positive)
Where f(x) is decreasing(Slope is negative)
Horizontal tangent (slope =0)
3.3 Differentiation formulas
Simple Power rule 1n ndx nx
dx
Sum and difference rule ( ) ( )d d d
u v u vdx dx dx
Constant multiple rule ( )d d
cu c udx dx
Constant ( ) 0d
cdx
Find the derivative function for:2 1
( ) 175 3
xf x
x
2( ) (3 2)f x x
12 21 1
( ) 175 3
f x x x
32
2 1( )
56
f x x
x
2( ) 9 12 4f x x x
( ) 18 12f x x
rewrite
rewrite
321 1 1
( ) 2* ( )( ) 05 3 2
f x x x
Rules for Finding Derivativesu and v are functions of x.
Simple Power rule 1n ndx nx
dx
Sum and difference rule ( ) ( )d d d
u v u vdx dx dx
Constant multiple rule ( )d d
cu c udx dx
Product rule ( ) ( )d d d
uv u v v udx dx dx
Quotient rule2
( ) ( )d d
v u u vd u dx dxdx v v
Differentiate2 1 3(3 2 )(4 5)y x x x
( ) ( )d d d
uv u v v udx dx dx
2 1 3 3 2 1(3 2 ) (4 5) (4 5) (3 2 )dy d d
x x x x x xdx dx dx
2 1 2 3 2(3 2 )(12 ) (4 5)(6 2 )dy
x x x x x xdx
4 4 2
4 2
36 24 24 8 30 10
60 14 10
dyx x x x x x
dxdy
x x xdx
Product rule
Differentiate 2
5 2
1
xy
x
2 2
2 2
( 1) (5 2) (5 2) ( 1)
( 1)
d dx x x xdy dx dx
dx x
2
( ) ( )d d
v u u vd u dx dxdx v v
2
2 2
( 1)(5) (5 2)(2 )
( 1)
dy x x x
dx x
2 2
2 2
(5 5) (10 4 )
( 1)
dy x x x
dx x
2
2 2
5 4 5
( 1)
dy x x
dx x
Quotient rule
Find the derivative function for:2( ) (3 2 )(5 4 )f x x x x
2
5 2( )
1
xf x
x
2 2( ) (3 2 ) (5 4 ) (5 4 ) (3 2 )d d
f x x x x x x xdx dx
2(3 2 )4 (5 4 )(3 4 )x x x x
224 4 15x x
2 2
2 2
( 1) (5 2) (5 2) ( 1)( )
( 1)
d dx x x x
dx dxf xx
2 2 2
2 2 2 2
2
2 2
( 1)5 (5 2)2 (5 5) (10 4 )
( 1) ( 1)
5 4 5
( 1)
x x x x x x
x
x x
xx
Velocity. The particle is moving forward for the first 3 seconds and backwards the next 2 sec, stands still for a second and then moves forward.
forward motion means velocity is positive
backward motion means velocity is negative
If velocity = 0, object is standing still.
The graphs of s and v as functions of time; s is largest when v = ds/dt = 0. The graph of s is not the path of the rock: It is a plot of height versus time. The slope of the plot is the rock’s velocity graphed here as a straight line.
a) How high does the rock go?
b) What is the velocity when the rock is 256 ft. above the ground on the way up? On the way down?c) What is the acceleration of the rock at any time?
d) When does the rock hit the ground? At what velocity?
3.4 applications
A dynamite blast blows a heavy rock straight up with a launch velocity of 160 ft/sec. Its height is given by s = -16t2 +160t.
A dynamite blast blows a heavy rock straight up with a launch velocity of 160 ft/sec. Its height is given by s = -16t2 +160t.
a) How high does the rock go?Maximum height occurs when v =0.
-32t + 160 = 0v = s´= -32t + 160
t = 5 sec.
s = -16t2 +160t
At t = 5, s = -16(5)2 +160(5) = 400 feet.
A dynamite blast blows a heavy rock straight up with a launch velocity of 160 ft/sec. Its height is given by s = -16t2 +160t.
b) What is the velocity when the rock is 256 ft. above the ground on the way up? On the way down?
v =-32t + 160at t = 2v=-32(2)+160 = 96 ft/sec.at t = 8v=-32(8)+160 = -96 ft/sec
-16t2 +160t = 256-16t2 +160t –256=0-16(t2 - 10t + 16)=0-16(t – 2) (t- 8) = 0t = 2 or t = 8
Find the times
Substitute the timesinto the velocityfunction
Set position = 256
A dynamite blast blows a heavy rock straight up with a launch velocity of 160 ft/sec. Its height is given by s = -16t2 +160t.
c) What is the acceleration of the rock at any time?
d) When does the rock hit the ground? At what velocity?
s = -16t2 +160t
v = s´= -32t + 160
a = v´ = s´´= -32ft/sec2
s = -16t2 +160t = 0t = 0 and t = 10
v =-32t + 160v = -32(10)+ 160 = -160 ft/sec.
Set position = 0
3.5 Derivatives of trig functions-formulas needed
sin(x+h) = sin x*cos h+cos x*sin h
0
sinhlim 1h h
cos(x+h) = cos x*cos h- sin x*sin h
0
cos 1lim 0h
h
h
Derivative of y = sin x0
( ) ( )( ) limh
dy f x h f xf x
dx h
0sin( ) sin( )
( ) limhx h x
f xh
0sin( )cos( ) cos( )sin( ) sin( )
limhx h x h x
h
0sin( )cos( ) sin( ) cos( )sin( )
limhx h x x h
h
0sin( )(cos( ) 1) cos( )sin( )
limhx h x h
h h
0sin( )(cos( ) 1) cos( )sin( )
limhx h x h
h h
0 +cos(x)*1 = cos (x)
3.5 Derivatives of Trigonometric Functions
sin cosd
x xdx
cos sind
x xdx
2tan secd
x xdx
2cot cscd
x xdx
sec sec tand
x x xdx
csc csc cotd
x x xdx
Figure 25: The curve y´ = –sin x as the graph of the slopes of the tangents to the curve y = cos x.
Slope of y = cos x
Find the derivatives21
( ) 5sin sec tan 7 32
f x x x x x x
21( ) 5cos sec tan sec tan (1) 14
2f x x x x x x x x
1 sin( )
cos
xf x
x x
2
( cos ) (1 sin ) (1 sin ) ( cos )( )
( cos )
d dx x x x x x
dx dxf xx x
2
( cos )(cos ) (1 sin )(1 sin )( )
( cos )
x x x x xf x
x x
2 2 2 2
2 2
( cos cos ) (1 sin ) cos cos 1 sin( )
( cos ) ( cos )
x x x x x x x xf x
x x x x
2
cos( )
( cos )
x xf x
x x