Ch 3 Derivatives, Antiderivatives,

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Ch 3 Derivatives, Antiderivatives,

    Ch 3: Derivatives, Antiderivatives, and Indefinite Integrals

    During the free-fall part of a sky diver’s descent, her downward acceleration is influenced by gravity and air resistance. The velocity increases at first, then approaches a limit called the terminal velocity. At any instant in time the acceleration is the derivative of the velocity. The distance she has fallen is the antiderivative, or indefinite integral, of the velocity. Velocity, acceleration, and displacement all vary, and depend on time.

    3.1 Graphical Interpretation of Derivative

     OBJ: Given the equation of a function, calculate the derivative at a given point by taking the limit of the quantity (change in y)/(change in x).

    3.2 Difference Quotients and One Definition of Derivative

    fxfc()()The derivative is approximately equal to This fraction is called the difference quotient.


    The derivative is the limit of the difference quotient as the denominator approaches zero.

     f’ (f prime) is the notation for derivative.

    DEFINITION: Derivative at a Point (derivative at x = c form)

    fxfc()()fc'()lim xc(xc

     Meaning: The instantaneous rate of change of f(x) with respect to x at x = c.

    OBJ: Given the equation of a function and a value of x, use the definition of derivative to calculate

    algebraically the value of the derivative at that point, and confirm your answer numerically and graphically.

    2EX: If f(x) = x 3x 4, find f’(5), the value of the derivative at x = 5. Check by graphing the difference

    quotient and finding the limit.

    Another way to check the answer is to graph the original function, then draw a line with slope 7 (the derivative) through the point (5, 6).

    Use y = mx + b to find the equation of the tangent line. The fact that this line is tangent to the graph is a graphical interpretation of the meaning of derivative.



    The derivative of a function at a point equals the slope of the tangent line to the graph of the function at that point. Both equal the instantaneous rate of change.

    3-3 Derivative Functions, Numerically and Graphically

    You have been calculating the derivative at a point x = c. Now it is time to find the derivative for all

    values of x.

    OBJ: Given the equation for a function, graph the function and its (numerical) derivative function on the same set of axes, and make conjectures about the relationship between the derivative function and the original function.

Graphers calculate derivatives using difference quotients.


    Often they use a symmetric difference quotient.


    fxhfx()();;fxfxh()();;You have used a backward or a forward difference quotient. or


    The x-increment, h, is often called the tolerance.

EX: Use a tolerance of h = 0.01 to estimate the forward, backward, and symmetric difference quotients of 3f(x) = x at x = 2.

     3Now use your grapher to find the numerical derivative of x with respect to x.

     3enter: nDeriv(x, x, 2)

Because there is a value of the derivative of each value of x, there is a function whose independent

    variable is x and whose dependent variable is the value of the derivative. The function is called the derivative function and is abbreviated f’(x). The grapher can plot this function by using an instruction such as y= nDeriv(y, x, x) 21

    32EX 1: Given f(x) = x 5x 8x + 70,

    a) Plot the graphs of f and f’ on the same screen.

b) f is a cubic function. What kind of function does f’ appear to be?

c) Trace to find f(3) and f’(x). Describe how f’(3) relates to the graph of f at x = 3.

    d) For what values of x does f’(x) = 0? What feature does the graph of f have at these x-values?

    e) Plot g(x) = f(x) + 10. On the same screen, plot g’. How is the graph of g related to the graph of f? How is the graph of g’ related to the graph of f’?

EX 2:

     Given f(x), sketch the graph of f’(x). Describe your process.

     32EX 3: Explore the cubic function f(x) = -x + 3x + 9x + 20 and its numerical derivative, f’. Make some conclusions about how certain features on the graph of f’, such as high and low points and x-intercepts, are related to features on the original function’s graph.

    3-4 Derivative of the Power Function and Another Definition of Derivative

     nOBJ: Given a power function, f(x) = x, where n stands for a constant, or given a linear combination of power functions, find an equation expressing f’(x) in terms of x.

The derivative of a function at a point x = c is

    fxfc()() fx'()limxc(xc

    This form of the definition is easy to remember because it ties in with the slope

    rise sloperun

    It also calls attention to the fact that you are finding the derivative at the one fixed point where x = c.

Another form of the definition leads more directly to an equation for a derivative. c is replaced by x and x

    is replaced by x + x.


DEFINITION: Derivative as a Function (x or h form)

    ?;?;;;yfxxfxfxhfx()()()() fx'()limlimlim????(?((xxh000??xxh

    Note: is the upper case of the Greek letter delta, while is the lower case of delta.

     5EX 1: Given f(x) = x, use the definition of derivative to find an equation for f’(x).

     PROPERTY: Derivative of the Power Function nn-1, then f’(x) =n x. If f(x) = x Restriction: The exponent n is a constant.

The process of finding an equation for the derivative of a function is called differentiation. The word

    reflects the fact that y/x is a difference quotient. The corresponding verb is differentiate.

PROCEDURE: Differentiating the Power Function nTo differentiate the power function, f(x)= x, multiply by the original exponent, n, then reduce the

    exponent by 1 to get the new exponent.

    THEOREM: If f(x) = g(x) + h(x), where g and h are differentiable functions, then f’(x) = g’(x) + h’(x) PROOF: By the definition of derivative,

    fxxfx()();?;fx'()lim ?(x0x

    gxxhxxgxhx()()()();?;;?;;,!,!lim = ?(x0x

    gxxgxhxxhx()()()();?;;;?;,!,! = lim?(x0x

    gxxgxhxxhx()()()();?;;?;~?lim; = ???(x0??xx?(

    gxxgxhxxhx()()()();?;;?;limlim; = ?(?(xx00??xx

     =g’(x) + h’(x)

    PROPERTIES: Differentiation

    Derivative of a Sum of Two Functions: If f(x) = g(x) + h(x), where g and h are differentiable functions of x, then f’(x) = g’(x) + h’(x).

Verbally: The derivative of a sum equals the sum of the derivatives. Differentiable distributes over


Derivative of a Constant Times a Function: If f(x) = kg(x), where g is a differentiable function of x,

    then f’(x) = kg’(x), provided k is a constant.

Verbally: The derivative of a constant times a function equals the constant times the derivative of the


    Derivative of a Constant Function: If f(x) = C, where C stands for a constant, then f’(x) = 0 for all values of x.

Verbally: Constants don’t change, so their rate of change is zero.

     73EX 2: If f(x) = 5x 11x + 12x -47, find f’(x).

    The answer is called

    the algebraic TERMINOLOGY: dy/dx and y’

    derivative. If y = f(x), then instead of writing f’(x) you can write any of the following.

     y’, read “y prime” (a short form of f’(x))


    , read “dy, dx” (a single symbol, not a fraction) dx

    d()y, read “d, dx, of y” (an operation done on y) dx

Regard dy/dx as a single symbol that cannot be taken apart avoid saying “dy over dx.”

    ddydWrite as where is the operator that tells you to take the derivative w.r.t. x, of y. ()ydxdxdx

    d54 ()5xxdx

     -3/4EX 3: If y = 6x , find dy/dx. Assume the power rule works for negative and rational exponents.

     5EX 4: If y = 2, find y’.

    3-5 Displacement, Velocity, and Acceleration

Obj: Given an equation for the displacement of a moving object, find an equation for its velocity and an

    equation for its acceleration, and use the equations to analyze the motion.

Suppose a football is punted into the air. As it rises and falls, its displacement (directed distance) from

    the ground is a function of the number of seconds since it was punted. 2 y = -16t + 37t + 37

    where y is the football’s displacement in feet and t is the number of seconds since it was punted.

The velocity of the ball gives its speed and the direction in which it’s going. Because velocity is the

    instantaneous rate of change, it is a derivative.

     velocity = dy/dt = y’ = -32t + 37

Find velocity at t = 1

     at t = 2

The dy/dt symbol reminds you of the units for velocity (ft/sec).

Speed is the absolute value of velocity. Speed tells how fast an object is going without regard to its


    Describe the speed and velocity at t = 1 and t = 2 sec.

    Note that the velocity changes from t = 1 to t = 2 sec. The instantaneous rate of change in velocity is called acceleration. Using v for velocity, v = -32t + 37.

    Find acceleration.

The dv/dt symbol for the derivative gives the units of acceleration. dv/dt is in (feet/second)/sec and written 2“ft/sec

Interpret the idea of negative acceleration.

     2Note that the acceleration is constant, -32 ft/sec, for an object acted on only by gravity.

To tell quickly whether an object is speeding up or slowing down, compare the signs of the velocity and


TECHNIQUE: Speeding Up or Slowing Down

     If velocity and acceleration have the same sign, the object is speeding up.

     If velocity and acceleration have different signs, the object is slowing down.

Acceleration is the derivative of a derivative, or the second derivative.

    The symbol for the second derivative of y w.r.t. t is

    2dy or f”(t) 2dt

EX 1: An object moves in the x-direction in such a way that its displacement from the y-axis is 32 x = 3t 30t + 64t + 57, for t 0

    where x is in miles and t is in hours. With your grapher in parametric mode, plot x as x and y = 1. Use 11

    path style, with [0, 10] t-step = .1 t

     [0, 200] and [0, 2] xy

    Write a paragraph describing the path. Include the times and places that the object reverses direction, and

    the time intervals during which the object is traveling to the right and to the left. A sketch may help.

EX 2: An object moves in the x-direction in such a way that its displacement from the y-axis is 32 x = 3t 30t + 64t + 57, for t 0 where x is in miles and t is in hours.

    a) Find equations for its velocity and acceleration.

b) Find the velocity and acceleration at t = 2, t = 4, and t = 6.

    At each time, state

     Whether x is increasing or decreasing, and at what rate.

     Whether the object is speeding up or slowing down, and how you decided

c) At what times in the interval [0, 8] is x at a maximum? Is x ever negative in the interval?

EX 3: The acceleration of a moving object is given by

    a(t) = 12t 5, where a(t) is in cm per min and t is in min. If the velocity at time t = 1 is 8 cm/min, find an

    equation for v(t), the velocity as a function of time. Use the equation to find the velocity at the instant

    t = 3.

This example is called an antiderivative or indefinite integral. The word indefinite is used because an

    antiderivative always has an unspecified constant, C, added. This constant is called the constant of

    integration. The velocity of a moving object is the antiderivative of the acceleration, and the

    displacement of the object is the antiderivative of the velocity.

     DEFINTION: Antiderivative, or Indefinite Integral Function g is an antiderivative (or indefinite integral) of function f iff g’(x) = f(x).

PROPERTIES: Velocity, Speed, and Acceleration

    If x is the displacement of a moving object from a fixed plane (such as the ground), and t is time, then

    Velocity: v = x’ = dx/dt

     22x/dt Acceleration: a = v’ = dv/dt = x’’ = d

Speed: |v|

    3-6 Introduction to Sine, Cosine, and

    Composite Functions

    Graph y = sin x and y = nder(sin x, x, x) TI 83/84/86

     y = d(sinx,x)|x=x TI 89

    Note that each graph is a sinusoid, and the derivative appears to be the cosine function.

    Graph y = cos x and y = nder(cos x, x, x) TI 83/84/86

     y = d(cos x,x)|x=x TI 89

    Note that the derivative appears to be the opposite of sin x.

    In Section 3-8, you will prove these properties=:)

     PROPERTIES: Derivatives of Sine and Cosine Functions

     dd(sin)cos (cos)sinxxxx??; dxdx

In this section you will try to discover how to find the derivative of a composite function, such as f(x) = 54sin(x) or g(x) = (cos x) where an operation is performed on the answer to another function.

    The function performed first is called the inside function (the fifth-power function in f and the cosine function in g, above). The function performed second, on the answer to the inside function, is called the

    outside function (the sine in f and the fourth power in g, above).

     OBJ: Work with study group to form conjectures on how to differentiate a composite function.

    3-7 Derivatives of Composite Functions-The Chain Rule

    22f(x) = sin (x) is called a composite function. It is composed of two other functions. The function x in

    parentheses is called the inside function. The sine function outside the parentheses is called the outside

    function. It operates on the answer to the first function.

    OBJ: Given the equation for a composite function, write the equation for its derivative function.

    2Suppose you want to differentiate y = sin (x).

    2Because the derivative of sine is cosine, you might think that y’ = cos(x)

    2Below is the graph y = sin (x) and its derivative.

    However, the amplitude is increases with the high points following the line y = 2x.

    2The derivative is y’ = 2x cos (x)

    2The anticipated result, cos (x), is multiplied by the derivative of the inside function, 2x. You can use the chain rule to differentiate composite functions.

    Verbally, the chain rule says to differentiate the outside function, while the inside function stays the same. Then multiply the result by the derivative of the inside function.

    Although derivatives are not fractions, you can use the definition of derivative to write them that way. (There is a complete description in the text page 103-4.)

    The name chain rule is used because when a composite function has several inside function, you get a whole “chain” of derivatives multiplied together.

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