Define each term or concept.

Chapter Differentiation Course Number Section.1 The Derivative an the Tangent Line Problem Objective: In this lesson you learne how to fin the erivative of a function using the limit efinition an unerstan the relationship between ifferentiability an continuity. Instructor Date Important Vocabulary Define each term or concept. Differentiation The process of fining the erivative of a function. Differentiable A function is ifferentiable at x if its erivative exists at x. I. The Tangent Line Problem (Pages 96 99) Essentially, the problem of fining the tangent line at a point P boils own to the problem of fining the slope of the tangent line at point P. You can approximate this slope using a secant line through the point of tangency (c, f(c)) an a secon point on the curve (c + Δx, f(c + Δx)). The slope of the secant line through these two points is m sec = f(c + Δx) f(c) Δx. How to fin the slope of the tangent line to a curve at a point The right sie of this equation for the slope of a secant line is calle a ifference quotient. The enominator Δx is the change in x, an the numerator Δ y = f( c+δx) f( c) is the change in y. The beauty of this proceure is that you can obtain more an more accurate approximations of the slope of the tangent line by choosing points closer an closer to the point of tangency. If f is efine on an open interval containing c, an if the limit Δ y f( c+δx) f( c) lim = lim = m exists, then the line passing Δx 0Δx Δx 0 Δx Larson/Ewars Calculus 9e Notetaking Guie IAE Copyright Cengage Learning. All rights reserve. 31

3 Chapter Differentiation through (c, f(c)) with slope m is the tangent line to the graph of f at the point (c, f(c)). The slope of the tangent line to the graph of f at the point (c, f(c)) is also calle the slope of the graph of f at x = c. x Example 1: Fin the slope of the graph of f( x ) = 9 at the point (4, 7). 1/ Example : Fin the slope of the graph of the point ( 1, 1). 6 f ( x) 3 = x at The efinition of a tangent line to a curve oes not cover the possibility of a vertical tangent line. If f is continuous at c an f c+δx f c lim Δ x 0 ( ) ( ) = or Δx f c+δx f c lim Δ x 0 ( ) ( ) =, the Δx vertical line x = c passing through (c, f(c)) is a vertical tangent line to the graph of f. II. The Derivative of a Function (Pages 99 101) The erivative of f at x is given by f '( x) = f x+δx f x lim Δ x 0 ( ) ( ), provie the limit exists. For all x Δx for which this limit exists, f is a function of x. How to use the limit efinition to fin the erivative of a function The erivative of a function of x gives the slope of the tangent line to the graph of f at the point (x, f(x)), provie that the graph has a tangent line at this point. A function is ifferentiable on an open interval (a, b) if it is ifferentiable at every point in the interval. Larson/Ewars Calculus 9e Notetaking Guie IAE Copyright Cengage Learning. All rights reserve.

Section.1 The Derivative an the Tangent Line Problem 33 Example 3: Fin the erivative of f (t) = 8t f t () 4t 5 = +. III. Differentiability an Continuity (Pages 101 103) Name some situations in which a function will not be ifferentiable at a point. A graph having a vertical tangent line or a graph with a sharp turn How to unerstan the relationship between ifferentiability an continuity If a function f is ifferentiable at x = c, then continuous at x = c. f is Complete the following statements. 1. If a function is ifferentiable at x = c, then it is continuous at x = c. So, ifferentiability implies continuity.. It is possible for a function to be continuous at x = c an not be ifferentiable at x = c. So, continuity oes not imply ifferentiability. Larson/Ewars Calculus 9e Notetaking Guie IAE Copyright Cengage Learning. All rights reserve.

34 Chapter Differentiation Aitional notes y y y x x x y y y x x x Homework Assignment Page(s) Exercises Larson/Ewars Calculus 9e Notetaking Guie IAE Copyright Cengage Learning. All rights reserve.

Section. Basic Differentiation an Rates of Change 35 Section. Basic Differentiation an Rates of Change Objective: In this lesson you learne how to fin the erivative of a function using basic ifferentiation rules. Course Number Instructor Date I. The Constant Rule (Page 107) The erivative of a constant function is zero. If c is a real number, then [] c = 0. II. The Power Rule (Pages 108 109) The Power Rule states that if n is a rational number, then the n function f ( x) = x is ifferentiable an x n = nx n 1. For f to be ifferentiable at How to fin the erivative of a function using the Constant Rule How to fin the erivative of a function using the Power Rule x = 0, n must be a number such that n 1 x is efine on an interval containing 0. Also, [ x ] = 1. 1 Example 1: Fin the erivative of the function f( x) =. 3 x 3/x 4 Example : Fin the slope of the graph of 80 f ( x) 5 = x at x =. III. The Constant Multiple Rule (Pages 110 111) The Constant Multiple Rule states that if f is a ifferentiable function an c is a real number then cf is also ifferentiable an [ cf ( x )] = cf (x). How to fin the erivative of a function using the Constant Multiple Rule Informally, the Constant Multiple Rule states that constants can be factore out of the ifferentiation process, even if the constants appear in the enominator. Larson/Ewars Calculus 9e Notetaking Guie IAE Copyright Cengage Learning. All rights reserve.

36 Chapter Differentiation Example 3: Fin the erivative of /5 x f( x ) = 5 The Constant Multiple Rule an the Power Rule can be combine into one rule. The combination rule is cx n = cnx n 1. Example 4: Fin the erivative of y = 5 5x /x 6 IV. The Sum an Difference Rules (Page 111) The Sum an Difference Rules of Differentiation state that the sum (or ifference) of two ifferentiable functions f an g is itself ifferentiable. Moreover, the erivative of f + g (or How to fin the erivative of a function using the Sum an Difference Rules f g ) is the sum (or ifference) of the erivatives of f an g. That is, [ f( x) g( x) ] an [ f( x) g( x) ] + = f (x) + g (x). = f (x) g (x). Example 5: Fin the erivative of 6x 8x +3 f x x x x 3 ( ) = 4 + 3 1 V. Derivatives of Sine an Cosine Functions (Page 11) [ sin x] = cos x How to fin the erivative of the sine function an of the cosine function [ cos x] = sin x Larson/Ewars Calculus 9e Notetaking Guie IAE Copyright Cengage Learning. All rights reserve.

Section. Basic Differentiation an Rates of Change 37 Example 6: Differentiate the function y' = x + sin x y x cosx =. VI. Rates of Change (Pages 113 114) The erivative can also be use to etermine the rate of change of one variable with respect to another. How to use erivatives to fin rates of change Give some examples of real-life applications of rates of change. Population growth rates, prouction rates, water flow rates, velocity, an acceleration. The function s that gives the position (relative to the origin) of an object as a function of time t is calle a position function. The average velocity of an object that is moving in a straight line is foun as follows. Average velocity = change in istance = change in time Example 7: If a ball is roppe from the top of a builing that is 00 feet tall, an air resistance is neglecte, the height s (in feet) of the ball at time t (in secons) is given by s= 16t + 00. Fin the average velocity of the object over the interval [1, 3]. 64 feet per secon Δs Δt If s = st ( ) is the position function for an object moving along a straight line, the (instantaneous) velocity of the object at time t is s(t + Δt) s(t) vt () = lim = s (t). Δt 0 Δt In other wors, the velocity function is the erivative of the position function. Velocity can be negative, zero, or Larson/Ewars Calculus 9e Notetaking Guie IAE Copyright Cengage Learning. All rights reserve.

38 Chapter Differentiation positive. The spee of an object is the absolute value of its velocity. Spee cannot be negative. Example 8: If a ball is roppe from the top of a builing that is 00 feet tall, an air resistance is neglecte, the height s (in feet) of the ball at time t (in secons) is given by st ( ) = 16t + 00. Fin the velocity of the ball when t = 3. 96 feet per secon The position function for a free-falling object (neglecting air resistance) uner the influence of gravity can be represente by the equation s(t) = 1/gt + v 0 t + s 0, where s 0 is the initial height of the object, v 0 is the initial velocity of the object, an g is the acceleration ue to gravity. On Earth, the value of g is approximately 3 feet per secon per secon or 9.8 meters per secon per secon. Homework Assignment Page(s) Exercises Larson/Ewars Calculus 9e Notetaking Guie IAE Copyright Cengage Learning. All rights reserve.

Section.3 Prouct an Quotient Rules an Higher-Orer Derivatives 39 Section.3 Prouct an Quotient Rules an Higher-Orer Derivatives Course Number Instructor Objective: In this lesson you learne how to fin the erivative of a function using the Prouct Rule an Quotient Rule. Date I. The Prouct Rule (Pages 119 10) The prouct of two ifferentiable functions f an g is itself ifferentiable. The Prouct Rule states that the erivative of the fg is equal to the first function times the erivative How to fin the erivative of a function using the Prouct Rule of the secon, plus the secon function times the erivative of the first. That is, [ f ( xgx ) ( )] = f ( xg ) ( x ) + gx ( ) f ( x ). Example 1: Fin the erivative of y/ = 4x 4x + y x x = (4 + 1)( 3). The Prouct Rule can be extene to cover proucts that have more than two factors. For example, if f, g, an h are ifferentiable functions of x, then [ f ( x ) g ( x ) h ( x )] = f (x)g(x)h(x) + f(x)g (x)h(x) + f(x)g(x)h (x) Explain the ifference between the Constant Multiple Rule an the Prouct Rule. The ifference between these two rules is that the Constant Multiple Rule eals with the prouct of a constant an a variable quantity, whereas the Prouct Rule Deals with the prouct of two variable quantities. Larson/Ewars Calculus 9e Notetaking Guie IAE Copyright Cengage Learning. All rights reserve.

40 Chapter Differentiation II. The Quotient Rule (Pages 11 13) The quotient f / g of two ifferentiable functions f an g is itself ifferentiable at all values of x for which gx ( ) 0. The How to fin the erivative of a function using the Quotient Rule erivative of f / g is given by the enominator times the erivative of the numerator minus the numerator times the erivative of the enominator, all ivie by the square of the enominator. This is calle the Quotient Rule, an is given by f( x) g( x) f ( x) f( x) g ( x) =, gx ( ) 0. g( x) gx ( ) [ ] Example : Fin the erivative of 5/(3x ) x + 5 y =. 3x With the Quotient Rule, it is a goo iea to enclose all factors an erivatives in parentheses an to pay special attention to the subtraction require in the numerator. III. Derivatives of Trigonometric Functions (Pages 13 14) [ tan x] = sec x How to fin the erivative of a trigonometric function [ cot x] = csc x [ sec x] = sec x tan x Larson/Ewars Calculus 9e Notetaking Guie IAE Copyright Cengage Learning. All rights reserve.

Section.3 Prouct an Quotient Rules an Higher-Orer Derivatives 41 [ csc x] = csc x cot x Example 3: Differentiate the function f ( x) = sin xsec x. f (x) = sin x sec x tan x + sec x cos x IV. Higher-Orer Derivatives (Page 15) The erivative of f ( x) is the secon erivative of f ( x ) an is enote by f (x). The erivative of f ( x) is the How to fin a higherorer erivative of a function thir erivative of f ( x ) an is enote by f. These are examples of higher-orer erivatives of f ( x ). The following notation is use to enote the sixth erivative. of the function y = f( x) : 6 y 6 D 6 [ y ] x (6) y 6 [ f ( x )] 6 f (6) ( x ) Example 4: Fin (5) y for 5040x 10 y x x 7 5 =. Example 5: On the moon, a ball is roppe from a height of 100 feet. Its height s (in feet) above the moon s 7 surface is given by s= t + 100. Fin the 10 height, the velocity, an the acceleration of the ball when t = 5 secons. Height: 3.5 feet above the surface Velocity: 7 feet per secon Acceleration: 7/5 feet per secon square Larson/Ewars Calculus 9e Notetaking Guie IAE Copyright Cengage Learning. All rights reserve.

4 Chapter Differentiation Example 6: Fin y for y = sin x. y = cos x Aitional notes Homework Assignment Page(s) Exercises Larson/Ewars Calculus 9e Notetaking Guie IAE Copyright Cengage Learning. All rights reserve.

Section.4 The Chain Rule 43 Section.4 The Chain Rule Objective: In this lesson you learne how to fin the erivative of a function using the Chain Rule an General Power Rule. Course Number Instructor Date I. The Chain Rule (Pages 130 13) The Chain Rule, one of the most powerful ifferentiation rules, eals with composite functions. How to fin the erivative of a composite function using the Chain Rule Basically, the Chain Rule states that if y changes y/u times as fast as u, an u changes u/ times as fast as x, then y changes (y/u)(u/) times as fast as x. The Chain Rule states that if y = f( u) is a ifferentiable function of u, an u = g( x) is a ifferentiable function of x, then y = f( g( x)) is a ifferentiable function of x, an y y = u or, equivalently, u [ f ( g ( x ))] = f (g(x))g (x). When applying the Chain Rule, it is helpful to think of the composite function f o g as having two parts, an inner part an an outer part. The Chain Rule tells you that the erivative of y = f( u) is the erivative of the outer function (at the inner function u) times the erivative of the inner function. That is, y = f (u) u. Example 1: Fin the erivative of 30x(3x ) 4 y 5 = (3x ). Larson/Ewars Calculus 9e Notetaking Guie IAE Copyright Cengage Learning. All rights reserve.

44 Chapter Differentiation II. The General Power Rule (Pages 13 133) The General Power Rule is a special case of the Chain Rule. The General Power Rule states that if y [ u x ] = ( ) n, where u is a How to fin the erivative of a function using the General Power Rule ifferentiable function of x an n is a rational number, then y = n[u(x)]n 1 u or, equivalently, u n = nu n 1 u. 4 Example : Fin the erivative of y = 3 (x 1) 4 (x 1) 4. III. Simplifying Derivatives (Page 134) 3x Example 3: Fin the erivative of y = (1 x ) y = 6x(x 3 + 1) / (1 x 3 ) 3 3 an simplify. How to simplify the erivative of a function using algebra Larson/Ewars Calculus 9e Notetaking Guie IAE Copyright Cengage Learning. All rights reserve.

Section.4 The Chain Rule 45 IV. Trigonometric Functions an the Chain Rule (Pages 135 136) Complete each of the following Chain Rule versions of the erivatives of the six trigonometric functions. How to fin the erivative of a trigonometric function using the Chain Rule [ sinu] = (cos u) u [ cosu] = (sin u) u [ tan u] = (sec u) u [ cot u] = (csc u) u [ secu] = (sec u tan u) u [ cscu] = (csc u cot u) u Example 4: Differentiate the function y/ = 4 sec 4x tan 4x y = sec4x. Larson/Ewars Calculus 9e Notetaking Guie IAE Copyright Cengage Learning. All rights reserve.

46 Chapter Differentiation Example 5: Differentiate the function y/ = x + sin(x + 1) y x x = cos( + 1). Aitional notes Homework Assignment Page(s) Exercises Larson/Ewars Calculus 9e Notetaking Guie IAE Copyright Cengage Learning. All rights reserve.

Section.5 Implicit Differentiation 47 Section.5 Implicit Differentiation Objective: In this lesson you learne how to fin the erivative of a function using implicit ifferentiation. Course Number Instructor Date I. Implicit an Explicit Functions (Page 141) Up to this point in the text, most functions have been expresse in explicit form y = f( x), meaning that the variable y is explicitly written as a function of x. However, some functions are only implie by an equation. How to istinguish between functions written in implicit form an explicit form Give an example of a function in which y is implicitly efine as a function of x. Answers will vary. For example, x y = 4 is in implicit form. Implicit ifferentiation is a proceure for taking the erivative of an implicit function when you are unable to solve for y as a function of x. To unerstan how to fin y implicitly, realize that the ifferentiation is taking place with respect to x. This means that when you ifferentiate terms involving x alone, you can ifferentiate as usual. However, when you ifferentiate terms involving y, you must apply the Chain Rule. because you are assuming that y is efine implicitly as a ifferentiable function of x. Example 1: Differentiate the expression with respect to x: 4x + y y 4 + y Larson/Ewars Calculus 9e Notetaking Guie IAE Copyright Cengage Learning. All rights reserve.

48 Chapter Differentiation II. Implicit Differentiation (Pages 14 145) Consier an equation involving x an y in which y is a ifferentiable function of x. List the four guielines for applying implicit ifferentiation to fin y/. How to use implicit ifferentiation to fin the erivative of a function 1. Differentiate both sies of the equation with respect to x.. Collect all terms involving y/ on the left sie of the equation an move all other terms to the right sie of the equation. 3. Factor y/ out of the left sie of the equation. 4. Solve for y/. Example : Fin y/ for the equation y/ = x/(4y) 4y x 1 =. Homework Assignment Page(s) Exercises Larson/Ewars Calculus 9e Notetaking Guie IAE Copyright Cengage Learning. All rights reserve.

Section.6 Relate Rates 49 Section.6 Relate Rates Objective: In this lesson you learne how to fin a relate rate. I. Fining Relate Variables (Page 149) Another important use of the Chain Rule is to fin the rates of change of two or more relate variables that are changing with respect to time. Course Number Instructor Date How to fin a relate rate Example 1: The variables x an y are ifferentiable functions of 3 t an are relate by the equation y = x x+ 4. When x =, /t = 1. Fin y/t when x =. 3 II. Problem Solving with Relate Rates (Pages 150 153) List the guielines for solving a relate-rate problems. 1. Ientify all given quantities an quantities to be etermine. Make a sketch an label the quantities. How to use relate rates to solve real-life problems. Write an equation involving the variables whose rates of change either are given or are to be etermine. 3. Using the Chain Rule, implicitly ifferentiate both sies of the equation with respect to time t. 4. After completing Step 3, substitute into the resulting equation all known values of the variables an their rates of change. Then solve for the require rate of change. Example : Write a mathematical moel for the following relate-rate problem situation: The population of a city is ecreasing at the rate of 100 people per month. x = number in population; /t = 100 people per month Larson/Ewars Calculus 9e Notetaking Guie IAE Copyright Cengage Learning. All rights reserve.

50 Chapter Differentiation Aitional notes Homework Assignment Page(s) Exercises Larson/Ewars Calculus 9e Notetaking Guie IAE Copyright Cengage Learning. All rights reserve.