aplikasi_pembezaan.ppt

Upload: noorazu

Post on 14-Apr-2018

220 views

Category:

Documents


0 download

TRANSCRIPT

  • 7/27/2019 aplikasi_pembezaan.ppt

    1/24

    The Derivative-Instantaneous rate of change

    0

    ( ) ( )( ) lim

    h

    dy f a h f af a

    dx h

    0

    ( ) ( )( ) limh

    dy f x h f xf x

    dx h

    The derivative of a function,fat a specific

    value ofx, say a is a value given by:

    The derivative of a function,fas a function of

    x, is calledf(x) and is given by:

  • 7/27/2019 aplikasi_pembezaan.ppt

    2/24

  • 7/27/2019 aplikasi_pembezaan.ppt

    3/24

    Related problems 2( ) 3 4 7f x x x

    1) Find the slope off(x) atx = 3,x = -2

    3) Find the point onf(x) for which the slope is 2

    4) Find the point for whichf(x) has a

    horizontal tangent line

    2) Write the equation of the tangent line at x = -2

  • 7/27/2019 aplikasi_pembezaan.ppt

    4/24

    Solutions2( ) 3 4 7f x x x

    1) Find the slope off(x) atx = 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

  • 7/27/2019 aplikasi_pembezaan.ppt

    5/24

    Solutions 2( ) 3 4 7f x x x

    3) Find the point onf(x) for which the slope is 2

    4) Find the point for whichf(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( ) 7

    3 3 3 3

    f

    The point is (2/3, 17/3)

  • 7/27/2019 aplikasi_pembezaan.ppt

    6/24

    Figure 2.7: Derivatives at endpoints are one-sided limits.

    Derivatives at Endpoints are one-sided limits.

  • 7/27/2019 aplikasi_pembezaan.ppt

    7/24

    How a derivative can fail to exist

    A cornerA vertical

    tangentA discontinuity

    Which of the three examples are the functions

    continuous?

  • 7/27/2019 aplikasi_pembezaan.ppt

    8/24

    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)

  • 7/27/2019 aplikasi_pembezaan.ppt

    9/24

    3.3 Differentiation formulas

    Simple Power rule 1n nd

    x nxdx

    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

  • 7/27/2019 aplikasi_pembezaan.ppt

    10/24

    Find the derivative function for:

    21

    ( ) 175 3

    xf x

    x

    2( ) (3 2)f x x

    12 2

    1 1( ) 17

    5 3

    f x x x

    3

    2

    2 1( )

    5

    6

    f x x

    x

    2( ) 9 12 4f x x x

    ( ) 18 12f x x

    rewrite

    rewrite

    3

    21 1 1

    ( ) 2 * ( )( ) 05 3 2f x x x

  • 7/27/2019 aplikasi_pembezaan.ppt

    11/24

    Rules for Finding Derivatives

    u and v are functions of x.

    Simple Power rule 1n nd x nxdx

    Sum and difference rule ( ) ( )d d du 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 dx

    dx v v

  • 7/27/2019 aplikasi_pembezaan.ppt

    12/24

    Differentiate

    2 1 3

    (3 2 )(4 5)y x x x

    ( ) ( )

    d d duv u v v u

    dx 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

    dx

    dyx x x

    dx

    Product rule

  • 7/27/2019 aplikasi_pembezaan.ppt

    13/24

    Differentiate 25 2

    1

    xy

    x

    2 2

    2 2

    ( 1) (5 2) (5 2) ( 1)

    ( 1)

    d dx x x x

    dy 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

  • 7/27/2019 aplikasi_pembezaan.ppt

    14/24

    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 df 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

  • 7/27/2019 aplikasi_pembezaan.ppt

    15/24

    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.

  • 7/27/2019 aplikasi_pembezaan.ppt

    16/24

    The graphs ofs and v as functions of time;s islargest when v = ds/dt= 0. The graph ofs isnotthe path of the rock: It is a plot of heightversus time. The slope of the plot is the rocks

    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.

  • 7/27/2019 aplikasi_pembezaan.ppt

    17/24

    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.

  • 7/27/2019 aplikasi_pembezaan.ppt

    18/24

    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 theground on the way up? On the way down?

    v =-32t+ 160

    at t= 2v=-32(2)+160 = 96 ft/sec.

    at t = 8

    v=-32(8)+160 = -96 ft/sec

    -16t2 +160t = 256

    -16t2 +160t256=0

    -16(t2 - 10t + 16)=0-16(t2) (t- 8) = 0

    t = 2 or t = 8

    Find the times

    Substitute the timesinto the velocity

    function

    Set position = 256

  • 7/27/2019 aplikasi_pembezaan.ppt

    19/24

    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 = 0

    t = 0 and t = 10

    v =-32t+ 160

    v = -32(10)+ 160 = -160 ft/sec.

    Set position = 0

  • 7/27/2019 aplikasi_pembezaan.ppt

    20/24

    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

  • 7/27/2019 aplikasi_pembezaan.ppt

    21/24

    Derivative of y = sin x0

    ( ) ( )( ) limh

    dy f x h f xf x

    dx h

    0

    sin( ) sin( )

    ( ) limh

    x h x

    f x h

    0

    sin( )cos( ) cos( )sin( ) sin( )limh

    x h x h x

    h

    0

    sin( )cos( ) sin( ) cos( )sin( )limh

    x h x x h

    h

    0

    sin( )(cos( ) 1) cos( )sin( )

    limhx h x h

    h h

    0

    sin( )(cos( ) 1) cos( )sin( )

    limhx h x h

    h h

    0 +cos(x)*1 = cos (x)

  • 7/27/2019 aplikasi_pembezaan.ppt

    22/24

    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

  • 7/27/2019 aplikasi_pembezaan.ppt

    23/24

    Figure 25: The curvey = sinx as the graph of the slopes of

    the tangents to the curvey = cosx.

    Slope of y = cos x

    Fi d h d i i

  • 7/27/2019 aplikasi_pembezaan.ppt

    24/24

    Find the derivatives

    21( ) 5sin sec tan 7 32

    f x x x x x x

    21( ) 5 cos sec tan sec tan (1) 142f 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 x

    x 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