Unit IV Derivatives 20 Hours Finish by Christmas
|
|
- Roland Stone
- 5 years ago
- Views:
Transcription
1 Unit IV Derivatives 20 Hours Finish by Christmas
2 Calculus There two main streams of Calculus: Differentiation Integration
3 Differentiation is used to find the rate of change of variables relative to one another. i.e. finding the slope of tangents lines Differentiation is also called taking the derivative of a function. Integration is used to find the accumulation of one quantity over time. i.e. Area under a curve
4 d From PHYSICS in Science 1206 Tangents to a displacement vs time curve gives the instantaneous velocity at a point What is the velocity at 3 seconds? t This is Differential Calculus
5 v 4 3 What did the area under a velocity time graph give you? 2 Displacement t What is the displacement after 3 seconds? This is Integral Calculus
6 In this unit of Derivatives we will find: Concept of the derivative. Derivative at a point. Derivative as a function. Computations of Derivative using definition of derivative Computations of Derivative using differentiation rules. Implicit Differentiation
7 Part A Introduction to Differential Calculus Demonstrate an understanding of the concept of a derivative and evaluate derivatives of functions using the definition of derivative.
8 The Details Explain what is meant by the term tangent. Calculate average rate of change. Identify the instantaneous rate of change of a function at a point as the limiting value of a sequence of average rates of change.
9 Derivatives: Tangents and Velocity First application of Calculus that we look at. What is meant by a tangent? Touches a curve at one point Circle Curve
10 Tangents in the Real World
11
12
13
14 For tangents we consider the line to touch at only a local level. We do not worry about cases like this Tangent here
15 Why are tangents important in Mathematics? A tangent gives an indication of the steepness of a curve at a given point. The slope of the tangent at a point gives the instantaneous rate of change of the function at that point. Consider the graph on the next page.
16 N P M The curve at M is steeper than the curve at P The curve is decreasing at N as compared to increasing at P and M We can get this information, without drawing the graph, by looking at the slopes of the tangent lines at these points.
17 Q R What is the local max point on this graph? Q What is the local min point on this graph? R What is the slope of the tangent lines at these extreme points? ZERO
18 Rate of Change
19 Definition: Rate of Change: The change in one quantity compared with the change in another quantity. Examples: Change in distance compared to time Speed Change in displacement compared to time Velocity Change in velocity compared to time Acceleration Change in temperature compared to altitude. Adiabatic lapse rate 2 o C 1000ft
20 Example: Lets look at a trip from Massey Drive to Bonne Bay Pond Time Location Total Distance 5:00 pm Massey Drive 5:10 pm Steady Brook 5:40 pm Deer Lake 0 km 10 km 50 km 5:51 pm Bonne Bay Pond 80 km What is the average speed for the complete trip? d 80km Average Speed 1hr t 51min 60min What is the average speed for each part of the trip?
21 Finding the average change in a function value (y) Consider the graph of y = x 2 4 Find the average change in y when x changes from 1 to 2. Find the slope of the secant line joining the 2 points y y 2 1 m x 2 x 1 0 ( 3) Therefore, the average change in y as x changes from 1 to 2 is 3
22 Note A line that connects 2 points on a curve is called a secant line.
23 In general, for any function y = f(x) The Average Rate Of Change in y (f(x)) for x changing from x 1 to x 2 is found using the slope formula: y y2 y1 f ( x 2) f x1 AROC x x x x x This formula provides the slope of the secant line between the points (x 1,f(x 1 )) and (x 2,f(x 2 )).
24 Example: For the function f(x) = -5x 2 +10x + 20 find the average rate of change in f(x) from x 1 = 0 to x 2 = 1
25 Instantaneous Rate of Change IROC How can you tell how fast you are going in a car? Look at the Speedometer What if your speedometer is broken, but your odometer still works? You could calculate your speed by finding the time it takes to go 1 km and divide 1km by that time.
26 How could you get a better idea how fast you are going right now, at this instant?? Find the time to go 0.1 km and divide 0.1km by that time. The idea is that by finding the average speed for smaller intervals you get a better idea at what your instantaneous speed is!
27 For graphs: The IROC for a graph is the SLOPE of a tangent drawn at a particular point For example: The IROC of y with respect to x at x = 1 is found by determining the slope of the tangent at the point (1,1)
28 Because it is difficult to accurately draw tangents we can approximate the slope of the tangent by finding the average slope of secant lines around the point in question
29 Here m sec > m tan Case I
30 Here m sec < m tan Case II
31 Lets consider Case I Where we take the point to the right of x = 1 Lets take a point close to x = 1, say x = 2. Draw the secant line. m sec y x f ( x ) x y x f x 2 1 x 2 1 NOTE : f ( x ) x 2 m sec We say that the slope of the tangent is close to 3
32 How can we get a better approximation? We take the point closer to x = 1 Lets use x 2 = 1.5. Draw the secant line. m sec f ( x2) f x1 x x 2 1 m sec We say that the slope of the tangent is close to 2.5
33 Is this approximation close Not Yet! enough to the slope?
34 Lets usex 2 = 1.1. Draw the secant f ( x ) msec x m sec f x 2 1 x We say that the slope of the tangent approximately 2.1 Or that the IROC of y with respect to x at x = 1 is 2.1
35 Not Yet! Is this approximation close enough to the slope? Where do we stop?? Well that is where Calculus comes in! We will find that the slope of the tangent line is the limiting value of the slopes of the secant lines as the secant line is drawn closer and closer to the tangent line.
36 Remember differentiation is used to find the rate of change of variables relative to one another. i.e. IROC which is represented graphically as the slope of tangents So how do we use calculus to find these slopes?
37 Finding slopes of tangent lines The first step to finding the slope of tangent lines is to make approximations using the slope of secant lines. Recall: A secant line is a line that connects 2 points on a curve.
38 Find the slope of the tangent line to the curve y = x 2 at the point P(2, 4) We find the approximate slope of the tangent by drawing lines (secant lines) that have approximately the same slope as the tangent line and calculate their slope.
39 Find the slope of the tangent line to the curve y = x 2 at the point P(2, 4) For the first approximation we draw line PQ with Q close to P, say (1,1). y2 y1 4 1 mpq 3 x x Thus we can say that the slope of the tangent line is close to 3.
40 How do we get a better approximation for the slope? We take the point Q to be closer and closer to P y2 y1 x Y m PQ x x x is approaching 2 x Thus we can say that the slope of the tangent line is close to 4.
41 We can take the point Q to be closer and closer to P from the right x Y m PQ y x y x Thus we can say that the slope of the tangent line is close to 4.
42 Thus we could say that as Q gets closer to P, the slope of the secant line gets closer to 4. From this we could guess that the slope of the tangent line at P is 4. We say that the slope of the tangent line is the limit of the slopes of the secant lines as Q approaches P.
43 Symbolically: lim m PQ QP m tan NOTE: m lim m PQ QP PQ Slope of secant lim y x x x y x x Slope of tangent 4 2 lim x 2 4 x 2 x 2 x lim x ( x 2)( x 2) x 2
44 What is the equation of the tangent line at P?
45 Tangent/Velocity Find the slope of a tangent line to the curve y = f(x) at point P (a, f(a)) f(x) f(a) a P x Q m m PQ PQ y x y x f ( x ) f ( a) x a And as x gets closer to a, the secant line gets closer to the tangent line.
46 In fact, the tangent line is the limiting line for the secant line as x gets closer to a. m lim m tan PQ QP m tan f ( x ) f ( a) lim xa x a
47 Definition: The tangent line to the curve y = f(x) at the point P(a, f(a)) is the line through P with slope f ( x ) f ( a) m lim xa x a provided the limit exists!
48 Examples 1. Find the equation of the tangent line to the curve y = x 2 at the point (- 1, 1). Use f ( x ) f ( a) the formula mtan lim. xa x a In this problem a = - 1 and f(x) = x 2.
49 2.Find the equation of the normal line to the graph y = x 2 + x at x = 1. NOTE: The normal line is perpendicular to the tangent. Find the slope of the tangent line first using the formula and then find the perpendicular slope. Use this to find the equation of the normal line.
50 3.Find the slope of the tangent to the 1 curve f(x) = at the point ( 0, 1) x 1
51 Velocity Example: What is the instantaneous rate of change (IROC) aka velocity at t = 2 s for an object traveling according the displacement function s(t) = -5t 2 +30t?
52 TEXT: Page 81 # , 10, 11, 15, 16, 18
53 New Definition: The definition of the tangent line can be changed into a different form by introducing a variable, h, to represent the distance that x is away from a. Consider the graph on the next page.
54 m tan a x SO, as x approaches a, h approaches zero thus taking the limit as is the same as taking the limit as h 0 h x a f ( x ) f ( a) lim xa x a Also h = x - a OR x = a + h Change the limit below such that x is removed
55 m m tan tan f ( x ) f ( a) lim xa x a f ( a h) f ( a) lim h h0
56 Apply this new formula we just derived to find the slope of the tangent line at the 1 point P ( 1, 1) for the function m tan f ( a h) f ( a) lim h0 h f( x) x
57 Find the equation of the tangent line 3 at the point for the function 2x 1 f( x) 3x 1 m tan 1, 2 f ( a h) f ( a) lim h0 h
58 The Velocity Problem
59 The Velocity Problem Suppose an object moves along a straight line according to an equation of motion s = f(t), where s is the displacement (directed distance) of the object from the origin at time t. The function f that describes the motion is called the position function of the object.
60 In the time interval from t = a to t = a + h the change in position is f (a + h) f (a). The average velocity over this time interval is: average velocity displacement f ( a h ) f ( a ) time h
61 From the graph of position and time we can see that the average velocity over this time interval is the same as the slope of the secant line PQ.
62 Now suppose we compute the average velocities over shorter and shorter time intervals [a, a + h]. In other words, we let h approach 0. We have defined the velocity (or instantaneous velocity) v (a) at time t = a to be the limit of these average velocities: va ( ) lim h0 f a h f a ( ) ( ) This gives the slope of the tangent line to the graph of s = f(t) when t = a. h
63 Example Suppose that a ball is dropped from the upper observation deck of the CN Tower, 450 m above the ground. (a) What is the velocity of the ball after 5 seconds? Use the equation of motion s = f (t) = 4.9t 2 to find the velocity v (a) after a seconds: f ( a h) f ( a) va ( ) lim h0 h
64 (b) How fast is the ball traveling when it hits the ground? Since the observation deck is 450 m above the ground, the ball will hit the ground at the time t 1 when s (t 1 ) = 450 The velocity of the ball as it hits the ground is therefore
65
66 # 12,13,14a) b) Text Page 81
67 Definition of Derivative Check up Each of the following represents the derivative of a function, f, at some number a. State f and a in each case. 1 1 ( i ) lim x 2 x 2 x 2 ( ii )lim h h 2 h ( iii )lim h 0 9h 3 h
68 In the Previous Section we: Defined and evaluated the derivative at x = a : m m tan tan f ( x ) f ( a) lim xa x a f ( a h) f ( a) lim h0 h Determined the equation of the tangent line and normal line to a graph of a relation at a given point. Used the derivative to find the velocity from position (displacement) function.
69 In the next section we will: Define and determine the derivative of a function using f ( x h) f ( x ) f( x) lim h0 h (limited to polynomials of degree 3, square root and rational functions with linear terms).
70 Use alternate notation interchangeably to express derivatives Determine whether a function is differentiable at a given point. Explain why a function is not differentiable at a given point, and distinguish between corners, cusps, discontinuities, and vertical tangents. Determine all values for which a function is differentiable, given the graph.
71 Sketch a graph of the derivative of a function, given the graph of a function. Sketch a graph of the function, given the graph of the derivative of a function.
72 Part B Derivatives Slope of a tangent line to the curve y = f(x) where x = a is given by: m tan f ( a h) f ( a) lim h0 h The velocity of an object with position s(t) at a time t = a is given by; s( a h) s( a) va ( ) lim h0 h
73 These two are basically the same equation. In fact, this type of equation is found whenever you calculate any rate of change in science, engineering, or business. It occurs so often that it is given a special name and notation
74 Definition of Derivative The derivative of a function f at x is denoted by f( x) and is expressed as: f ( x h) f ( x ) f( x) lim h0 h provided the limit exists!!
75 Domain of the derivative function. The domain of the new function f( x) is the set of all numbers x for which the limit exists. Also the domain of f( x) is a subset of the domain of f(x) because f(x) is used in the definition of f( x)
76 Examples: 1. Find the derivative of the function f(x) = x f ( x h) f ( x ) f( x) lim h0 h
77 Examples: 2. Find the derivative of the function f(x) = x 3 x f ( x h) f ( x ) f( x) lim h0 h
78 3. Find g ( x) for g( x) lim h0 Examples: 1 gx ( ) x 2 State the domain of g(x) and g( x h) g( x ) h g( x)
79 Examples: ( ) x 1 4. Find g ( x) for g x State the domain of g(x) and g( x) lim h0 g( x h) g( x ) h g( x)
80
81 Examples: 5. Differentiate (Find f ( x) ) f ( x h) f ( x ) f( x) lim h0 h f( x) 1 2 x
82
83 Find s() t CRAZY ONE for st () 1 2t 3t 4
84 Text Page 82 #31-36 Page 93 #19, 21, 23, 25, 27
85 Notes on Notation The derivative of a function y = f(x) at x is: f ( x h) f ( x ) f( x) lim h0 The derivative can also be written as: f( x) y OR AS h dy df dx d f dx ( x dx ) Leibniz Notation means the derivative of y (or f) with respect to x D f ( x ) D f ( x ) x
86 dy dx Leibniz Notation should not be regarded as a ratio. It is just another way of saying The advantage of the Leibniz notation is that both the independent variable (x) and the dependent variable (y) are shown in the derivative Example: Velocity v ( t ) s( x ) f( x) ds dt
87 Evaluating a Derivative To evaluate a derivative of y = f(x) at a certain value x = a we could say: f( a) or in Leibniz notation we would say: dy dx x a
88 d and D x are called differentiation dx operators. Differentiation is the process of calculating a derivative A function is differentiable at x = a if f( a) exists. A function is differentiable over an interval if it is differentiable at every number in the interval.
89 Theorem If a function f is differentiable at x = a then f is continuous at x = a. This means differentiable continuous but continuous differentiable However not continuous not differentiable
90 If a function is continuous at x = a, its graph is uninterrupted at x = a. If a function is differentiable at x = a, its graph is uninterrupted and smooth at x = a. Smooth means no sharp points. Also even if the function is smooth and continuous at x = a it would not be differentiable the if the tangent was a vertical line. (Infinite slope)
91 Examples: Discuss the differentiability of the following: This graph is 1. y = x 2 continuous and differentiable everywhere. What is y? y This is defined everywhere. 2x
92 2. If a function is NOT continuous at x = a then it is NOT differentiable a Note: f(x) is differentiable for all other values other than x = a
93 3. Is g(x) differentiable at x = 1? NO! g(x) is NOT continuous at x = 1 thus its is NOT differentiable at x = 1
94 4. y = x This function is continuous for but it is not differentiable at (0,0) There is a sharp point, (or corner point) at x = 0. With no unique tangent the function is not differentiable at x = 0,
95 Lets find the derivative of f(x) = x at x = 0 f (0 h) f (0) 0 h 0 f (0) lim lim h0 h h0 h h lim undefined h 0 h h lim h h lim 0 h h 0 h0 h 0 h lim 1 h h lim 1 h h Therefore the derivative does not exist at x = 0
96 5. y = f(x) This function is continuous, but it is not differentiable at (1,3) There is a sharp point at x = 1. There are two tangent lines depending on whether you approach x= 1 from left or right With no unique tangent the function is not differentiable at x = 1
97 2 6. Sketch the graph of (use technology) y y x 3 This function is continuous in the interval x but it is not differentiable at (0,0) There is a cusp at x = 0 A cusp is an extreme case of a corner point where the slopes of the secant lines approach from one side of the curve and - from the other. With no unique tangent the function and the fact that the tangent has an infinite slope the function is not differentiable at x = 0,
98 7. y = g(x) Is g(x) differentiable at x = 1? NO! g(x) is continuous at x = 1 but it is NOT differentiable at x = 1 because f(x) has a vertical tangent at x = 1. The tangent has an undefined slope. (Infinite slope)
99 In general, Functions whose graphs have : corners Cusps sharp points Vertical tangents are not differentiable at these points
100 Text Page 94 #33-37
101 Differentiability Check- UP 1. Consider the function f( x) 2 x, x 1 2 x 2, x 1 A) Is the function continuous at x = 1 B) Is the function differentiable at x = 1?
102
103 Is there a cusp, vertical tangent, discontinuity or corner point on the graph at x = 1?
104 2. Find all the points on the graph where the function is not differentiable. Determine why the function is not differentiable at these points.
105 Derivative as a Function Here we construct the graph of the derivative of f(x) from the graph of y = f(x) and vice versa. REMEMBER: When the graph of f (x) has a positive slope, the graph of fx will be above the x-axis. When the graph of f (x) has zero slope the graph of fx will cross the x-axis. When the graph of f (x) has a negative slope, the graph of f x will be below the x-axis.
106 Sketch the graph of y = f (x) given the graph of y = f(x) 1. f(x) = x 2 y = f (x)
107 Sketch the graph of y = f (x) given the graph of y = f(x) 2. y = f(x) y = f (x)
108 Note: For polynomials the degree of the derivative is always one less (one less bump) than the degree of the original polynomial
109 Sketch the graph of y = f (x) given the graph of y = f(x) 3. y = f(x) y = f (x)
110 Sketch the graph of y = f (x) given the graph of y = f(x) 4. y = f(x) y = f (x)
111 Sketch the graph of y = f (x) given the graph of y = f(x) 5. y = f(x) y = f (x)
112 Sketch the graph of y = f (x) given the graph of y y = f(x) 6. x
113 Practice
114 Text Page 92 #3, 4, 9, 11
115 Construct the graph of f(x) given fx y f ( 1) 2 x
116 2. Construct the graph of f(x) given fx y f (1) 3 x
117 3. Construct the graph of f(x) given fx y x
118 Part B C4. Apply derivative rules including: Constant Rule Constant Multiple Rule Sum Rule Difference Rule Product Rule Quotient Rule Power Rule Chain Rule
119 Details We will derive the Constant Rule, Sum and Difference Rules, Product and Quotient Rules but these proofs are not required for assessment. We will determine derivatives of functions, using the Constant, Constant Multiple, Power, Sum, Difference, Product and Quotient Rules.
120 We will determine second and higher-order derivatives of functions. We will determine derivatives of functions using the Chain Rule. We will solve problems involving derivatives drawn from a variety of applications, limited to tangent and normal lines, straight line motion and rates of change.
121 Computation of derivatives.
122 Rules for Differentiation Constant Rule: If f(x) = c, where c is a constant, then f( x) 0 or d c 0 dx Graphically: Consider y = 2 What is the slope of the tangent? 0
123 Proof: f ( x h) f ( x ) f( x) lim h0 h c c 0 lim lim h0 h h 0 h lim 0 0 h0 Example: Find the derivative of: A) f(x) = 5 B) y = -3 y C) D) y = 0
124 General Power Rule If n is any REAL number then d x dx n nx n1
125 Examples: 1. 5 A) If f ( x ) x then f( x) 100 B) If y x then y
126 Examples: 6 C) If y t then dy dt D) d r 3 dr
127 Examples: 2. Differentiate: A) 1 f( x) 3 x We must rewrite the function as a power of x B) f ( x ) x
128 Examples: 2. Differentiate: C) f ( x ) 3 x 2 D) f ( x ) x
129 Note: The Power Rule enables us to find slopes of tangent lines without having to resort to the definition of a derivative.
130
131
132 Constant Rule (II) (Or the Constant Multiplier Rule) If g(x) = c f(x) then g x where c is a constant PROOF: ( ) cf ( x ) g( x h) g( x ) g( x) lim h0 h cf ( x h) cf ( x ) g( x) lim h0 h f ( x h) f ( x ) c lim h0 h cf ( x )
133 In Leibniz notation: d dx d dx cf x c f x
134 Example: Differentiate: 3 A) f ( x ) 2x B) f ( x ) 4 x 3 C) f( x) 12x 3 1 4
135 Text Page 105 # 1, 2, 4, 11, 12, 18
136 Sum and Difference Rule The next rule tells us that the derivative of a sum (or difference) of functions is the sum (or difference) of the derivatives. If f(x) and g(x) are both differentiable functions then so is f ( x ) g( x ) AND f ( x ) g( x ) f ( x ) g ( x ) OR d f ( x ) g( x ) d f ( x ) d g( x ) dx dx dx
137 Proof: (of difference rule) Let k(x) = f(x) g(x) k ( x h) k ( x ) k( x) lim h0 h f ( x h) g( x h) ( f ( x ) g( x )) k( x) lim h0 h f ( x h) g( x h) f ( x ) g( x ) lim h0 h f ( x h) f ( x ) g( x h) g( x ) lim h0 h
138 f ( x h) f ( x ) lim h lim h0 h0 g( x h) g( x ) h f ( x ) g( x )
139 Note: This rule can be extended to any number of functions f ( x ) g( x ) k ( x ) s ( x ) f ( x ) g( x ) k ( x ) s( x )
140 Examples 1.
141 d dx 2. x 2 ( x 1)
142 d dx 3. 2x 3 3
143 4. dx 4 d x x x 2
144 5. dx 3 2 d x x x 4 x
145 Text Page 105 # 3, 5, 6, 10, 16, 19, 26
146 Product Rule The following formula was discovered by Leibniz and is called the Product Rule. If P(x) = f(x)g(x) and both f(x) and g(x) are differentiable then: d f ( x ) g( x ) d f x g x f x d g x dx dx dx OR ( ) ( ) ( ) ( ) ( ) P x f x g x f x g x
147 In words, the Product Rule says that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.
148 PROOF: P ( x h) P ( x ) P( x) lim h0 h f ( x h) g( x h) f ( x ) g( x ) P( x) lim h0 h Math Magic Introduce : f ( x h) g( x ) OR g( x h) f ( x ) f ( x h) g( x h) f ( x h) g( x ) f ( x h) g( x ) f ( x ) g( x ) lim h0 h f ( x h)[ g( x h) g( x )] g( x )[ f ( x h) f ( x )] lim h0 h
149 f ( x h)[ g( x h) g( x )] g( x )[ f ( x h) f ( x )] lim lim h h h0 h0 [ g( x h) g( x )] [ f ( x h) f ( x )] lim f ( x h)lim g( x )lim h0 h0 h h0 h P ( x ) f ( x ) g( x ) g( x ) f ( x ) OR P ( x ) f ( x ) g( x ) f ( x ) g( x )
150 Examples Differentiate: A) P(x) = (3x + 2)(4x 3) OLD WAY Expand P(X) = 12x 2 x 6 P ( x ) 24x 1 Product Rule: P ( x ) (3x 2) (4x 3) (3x 2)(4x 3) 3(4x 3) (3x 2)4 12x 9 12x 8 24x 1
151 NOTE: The product rule is more useful for finding the derivative of functions like: y = x 2 sinx y x x 1 x y e cosx However we need to have the transcendental rules first!
152 Examples: Differentiate B) 2 y x x 2
153 C) Suppose f(2) = 5, f (2) = 6, g(2) = 7 and g (2) = -1, find (fg) (2) ( fg) (2) f (2) g(2) f (2) g(2)
154 D) If h(x) = xg(x) and it is known that g(3) = 5 and g(3) = 2, find h(3).
155
156 Quotient Rule If Qx ( ) f ( x ) gx ( ) and both f(x) and g(x) are differentiable then: f ( x ) g( x ) f ( x ) g ( x ) Q ( x) 2 gx ( )
157 In words, the Quotient Rule says that the derivative of a quotient is the derivative of the numerator times the denominator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.
158 PROOF: Use the product rule! Qx ( ) f( x) f ( x ) Q( x ) g( x ) gx ( ) ( ) ( ) ( ) ( ) ( ) f x Q x g x Q x g x Solve for Q( x) ( ) ( ) ( ) ( ) ( ) Q x g x f x Q x g x f( x) Q( x ) g( x ) f ( x ) g( x ) gx ( )
159 gx ( ) f( x) Q ( x ) g( x ) f ( x ) g ( x ) g( x ) g( x ) Q( x ) g( x ) Q( x) f ( x ) g( x ) f ( x ) g( x ) gx ( ) f ( x ) g( x ) f ( x ) g( x ) gx ( ) 2 This proof can also be done using the definition of derivative in the same way that the product rule was proven!
160 Examples: Differentiate. x 2 A) y x 1
161 B) y 3x 1 2x 5
162 C) y 2x x 2 2 x 1
163 D) y 6 1 x x
164
165
166 Higher Derivatives
167 Higher Derivatives If f is a differentiable function, then its derivative f is also a function, so f may have a derivative of its own, denoted by (f) = f. This new function f is called the second derivative of f because it is the derivative of the derivative of f.
168 The Second Derivative d f ( x ) f ( x ) dx f ( x h) f ( x ) f( x) lim h0 h : Using Leibniz notation, we write the second derivative of as y = f (x) as : d d f ( x ) f ( x ) dx dx 2 d ( ) 2 f x dx 2 d y dx 2
169 Find the 2 nd derivative of: 1. f(x) = x 2 + 2x
170 2. f(x) = x 3 3x
171 3. s(t) = -4.9t t + 200
172 Higher Derivatives The 2 nd derivative has applications in curve sketching and physics. In general, we can interpret a second derivative as a rate of change of a rate of change. The most familiar example of this is acceleration. The instantaneous rate of change of velocity with respect to time is called the acceleration of the object.
173 Example A robot moves along a straight line according to the function: s t t t t 3 2 ( ) A) Determine when the robot is stopped. B) Determine the acceleration when the robot is stopped
174
175 Practice: Text Page 105 #43
176 In curve sketching the 2 nd derivative tells you how fast the slopes of the tangents are changing. This is called concavity.
177 Example: 1. y = x 2 Find y y = 2x y = 2 The fact that the 2 nd derivative is positive tells you that the tangent lines are increasing. It also tells you that the graph is concave upwards. Holds water Or that all of the tangent lines lie below the curve
178 Example: 2. y = -x 2 3x +4 Find y y = -2x - 3 y = -2 The fact that the 2 nd derivative is negative tells you that the tangent lines are decreasing. It also tells you that the graph is concave downwards. Does NOT Hold water Or that all of the tangent lines lie above the curve
179 Sketch the graph of y = f (x) and y = f (x) given the graph of y = f(x) y = f(x) y = f (x) y = f (x)
180 Sketch the graph of y = f (x) and y = f (x) if f(x) = sin x y = f(x) y = f (x) y = f (x) y y y x x x
181 Higher Derivatives The third derivative f is the derivative of the second derivative f ( x ) f x So y f ( x ) can be interpreted as the slope of the curve y = f (x) or as the rate of change of f (x).
182 Higher Derivatives If y = f(x), then alternative notations for the third derivative are d 2 d y d 3 y 2 dx dx dx 3 The process can be continued but the multiple prime notation will eventually become cumbersome for higher derivatives. y f x The fourth derivative f is usually denoted by f (4).
183 In general, the nth derivative of f is denoted f (n) by and is obtained from f by differentiating n times. If y = f(x), we write n y n f x n d y dx n
184 Note: The superscripts used for higherorder derivatives, such as y (4) for the fourth derivative of y, are not exponents. The presence of the parenthesis in the exponent denotes differentiation 2 (i.e., f x f x ) while the absence of the parenthesis denotes exponentiation (i.e., f 2 x f x 2 ).
185 Examples 1.Determine the first five derivatives of f ( x ) x 2x 2x 3x 4
186 Given f ( x ) 3x 2x 7x 9 determine the value of n, such that n 0 f x
187 3.Determine the point on the graph of 3 2 y 2x 12x 5x 6 where 2 d y 0 2 dx
188 DERIVATIVES OF COMPOSITE FUNCTIONS CHAIN RULE IMPLICIT DIFFERENTATION
189 Chain Rule Differentiate: A) (x + 1) 2
190 B) (x + 2)3
191 C) (x2 + 1)3
192 Lets break C) into a composition of functions f(g(x)) What would be the outer function f(x)? What would be the inner function g(x)? Therefore f(g(x)) = f(x 2 + 1)= (x 2 + 1) 3 And f ( x ) Which means x 3 g( x ) x 2 1 d f ( g ( x )) dx 3x x d f ( g ( x )) f ( g ( x )) g ( x ) dx
193 Basically the Chain Rule is a way of finding the derivative of a composition of functions Chain Rule: If C ( x ) f g( x ) f ( g( x )) C ( x ) f ( g( x )) g( x ) then derivative of the outer function wrt the inner function derivative of the inner function
194 Lets consider derivatives as IROC Example: A car travels twice as fast as a bicycle A bicycle travels 4 times as fast as a person walking A car travels 8 times as fast as a person walking Numerical example: A person walks at 5 km/h The bicycle travels at 20 km/h The car travels at 40 km/h 8 times as fast
195 This can be applied to any situation involving related IROCS dy du If y changes a times as fast as u Outer function du dx Inner function and u changes b times as fast as x then y changes ab times as fast as x dy dx ab dy dy du dx du dx Leibniz notation for the Chain Rule
196 Referencing back to C) y = (x 2 + 1) 3 how can the Leibniz notation for the Chain rule be applied to this problem? What is the inner function for y? x 2 +1 The function u is the inner function Rewrite y = (x 2 + 1) 3 using u y = (u) 3 Differentiate using dy dy dx dy du du dx d du dy du dx du dx d dx 3 2 u x 1 2 3u 2x 2 x 3 1 2x
197 Examples: 1.Find A) d dx x 1 2 B) d dx 2x 1 3 C) d dx x
198 D) d dx x 2 x E) d dx x 2x 5 4
199 2.Find A) d 2 dx x x
200 d dx 3 B) 4x 4x
201 3.A) If y u u , where u x 3 find dy dx
202 3.B) If y = u 10 7u 3 + 1, where u = 3x 2 2x, find dy dx
203 4. If h(x) = f(g(x)), find h(2) when g(2) = 3, g(2) 5, f (2) 1, f (3) 2 and f (5) 4
204 Text Page # 1, 2, 7, 8, 9, 11, 53, 54,55, 56
205 Fun Examples Differentiate the following functions: 5 A) f ( x ) 2x 1 4x 1
206 B) y 2x 3 3 4x 7
207 3 C ) f ( x ) 2x 3x
208 D) f ( t ) 2 t 1 t 1 10
209 dy 2. Determine the value of where dy 5, dv 2 x and du 3 du dx dv dx
210
211
212
213
214 Solve problems involving derivatives drawn from a variety of applications limited to tangent and normal lines, straight line motion and rates of change.
215 Tangent and Normal Lines 1. Find the equation of the tangent 1 line to the curve y at the x point 2, 2
216 2. Determine the equation of the tangent line to the curve at the given value. 3 2 A) y x x x 1, at x 1
217 3 2 B) y x 2, at x 2
218 C) y 7 x 3 x, at x 1
219 3. Find the equation of the normal line to the curve 3 s( t ) t 2t 5, at x 11
220 4. At what point on the curve 4 y x x 25 2 is the tangent parallel to the line 7x y 2?
221 5. Find the equation of both tangent lines that pass through the origin and are tangent to the parabola y = x (0, 0) (x, x 2 + 1) How do we find the slope of the tangent line? y2 y1 mtan x x x 10 f( x) x 0 2 x 1 2x x 2 2 2x x 1 2 x 1 0 x 1
222 Thus the slopes of the tangents are: For x = -1 m tan = 2x = 2(-1) = -2 For x = 1 m tan = 2x = 2(1) = 2 And the equation is: (y 0) = -2(x 0) y = -2x And the equation is: (y 0) = 2(x 0) y = 2x
223 2 y x x 6. (YOUR TURN) Find the equations of the 2 tangent lines to the parabola y x x that pass through the point 2, 3. Sketch the curve and the tangents
224 y x
225 7. Find the x-coordinates of the points on the hyperbola xy 1 where the tangents from the point 1, 1 intersect the curve.
226 Text Page #17, 21, 26, 48, 75
227 Straight Line Motion velocity (rate of change of displacement with respect to time) If an object travels along a path, and displacement s(t) is measured over time, then the first derivative s'(t), equals the velocity v(t) of that object. (rate acceleration of change of velocity with respect to time) The second derivative s"(t) = v'(t) equals the acceleration a(t) of that object.
228 For Straight Line Motion Problems, when v(t) = 0 object is not moving v(t) > 0 (positive) object is moving forward (to the right) v(t) < 0 (negative) object is moving backward (to the left)
229 For Straight Line Motion Problems, when no change in the velocity a(t) = 0 there is a(t) > 0 (positive) object s velocity is increasing a(t) < 0 (negative) object s velocity is decreasing
230 NOTE: If the signs of v(t) and a(t) are the same (both positive or both negative),then the speed of the object is increasing If the signs of v(t) and a(t) are opposite (one positive and the other negative), then the speed of the object is decreasing
231 Example A particle moves back and forth along a horizontal line defined by the position function: s(t) = t 3-12t t - 30, t 0. (a) Determine the velocity and acceleration functions. 2 v( t ) s t 3t 24t 36 a( t ) v t 6t 24 (b) When is the velocity zero? When is the acceleration zero? vt ( ) 0 at ( ) 0 2 3t 24t ( t 8t 12) 0 3 t 2 t 6 0 t 2or t 6 6t 24 0 t 4
232 (c) During what time intervals is the velocity positive? Explain what it means. vt ( ) 0 t t [0,2) (6, ) The particle is travelling to the right from t = 0 seconds to t = 2 seconds and after 6 seconds (d) During what time interval is the velocity negative? Explain what it means. vt ( ) 0 t t 2, The particle is travelling to the left from t = 2 seconds to t = 6 seconds
233 (e) When is the particle speeding up? Slowing down? This depends on the acceleration of the particle and the direction it is travelling. at ( ) 0 6t t 24 and at ( ) 0 t 4 t 4 (f ) Provide an overall description of what is happening to the particle. So for the first 2 seconds, the particle is travelling right but slowing down. At 2 seconds it stops and then speeds up in as it travels left. At 4 seconds the acceleration is zero and the particle has reached its max speed in the left direction.
234 After 4 seconds the acceleration is positive and thus the particle slows down and stops travelling left at 6 seconds. After 6 seconds the particle travels right and speeds up.
235 (g) What is the total distance travelled by the particle during the first minute?
236 Text Page #43, 44 a) b), 50 Page 142 # 65
237 Warm UP A demographer develops the function P(x) = x x 3 to represent the population of the town Tanville x years from now. (a) Determine the population of the town today. (b) Predict the instantaneous rate of change of the population during year 3.
238 Warm UP P(x) = x x 3 (c) Determine the average rate of change in the population between years 2 and 7. (d) In what year can you expect the population to increase by 245 people? (e) Is the rate of change in the population increasing or decreasing? Explain.
239 Part C C5. Determine the derivative of a relation, using implicit differentiation. Determine the derivative of an implicit relation. Determine the equation of the tangent and normal line to the graph of a relation at a given point. Determine the second derivative of a relation, using implicit differentiation.
240 Implicit Differentiation Function y = 2 y = c y = x y = x 2 y = (5x+1) 2 y = u 2 Explicitly Defined Derivative dy 0 dx dy 0 dx dy 1 dx dy 2x dx y y 2 5x 1 5 2u u x = y 2 This is an implicitly defined relation. 1 2yy
241 NOTE: x = y 2 can be solved explicitly and then differentiated.
242 One of the first implicit relations you encountered was the equation of a circle: x 2 + y 2 = r 2 Suppose you are asked to find the slope of the tangent line to the circle centred at the origin with a radius of 10 at the point (6, 8) You would need to differentiate the equation of the circle and substitute x = 6 into the derivative. What is the equation for this circle?
243 Lets find dy dx for x 2 + y 2 = 10 2 both explicitly and implicitly y 100 x 2 x y dy dx 2 dy x dx x x 2 x dy 2x 2y 0 dx dy 2y 2x dx dy 2x x dx 2y y Sub in the point (6, 8) dy dx dy dx dy dx
244 Note: The derivative expression for explicit differentiation involves x only, while the derivative expression for implicit differentiation may involve both x and y. If you have a choice you can solve by implicit differentiation or by first solving the given equation for y, however you are encouraged to choose the most efficient method.
245 Note: There are, however, examples where it is impossible to solve the equation for y as an explicit function of x. In these cases, the method of implicit differentiation becomes the only choice for solving for y'.
246 Examples: 1. Find y A) xy = 1
247 B) x 3 + 2x + y + y 3 = 3
248 C) x 2 + xy + y 2 = 5
249 2. Find for 2x 3 u x u 2 = u x 2 du dx
250 3. Find the equation of the A) tangent and B) normal lines to the curve 3x 2-2xy + xy 3 = 7 at the point (1, 2)
251 4. Find y A) x 2 y 2 y 3 = xy
252 B) x y xy 6
253
254 5. PROVE: when x 3 + y 3 = 10 y 20x 5 y
255 6. Prove that the tangent to a circle at a point P is perpendicular to the radius drawn to P m m (a,b) C r P (x,y) Show y mcp x tan y x CP y b x a The perpendicular slope is: m CP This is what should dx equal dy x a y b Equation of circle is: (x a) 2 + (y - b) 2 = r 2 dy Find dx
256
257 7. The curve 2(x 2 + y 2 ) 2 = 25(x 2 - y 2 ) shown below is called a lemniscate. Determine the equation of the tangent line to this curve at (3,1).
258
259
260
Unit IV Derivatives 20 Hours Finish by Christmas
Unit IV Derivatives 20 Hours Finish by Christmas Calculus There two main streams of Calculus: Differentiation Integration Differentiation is used to find the rate of change of variables relative to one
More informationCalculus I. 1. Limits and Continuity
2301107 Calculus I 1. Limits and Continuity Outline 1.1. Limits 1.1.1 Motivation:Tangent 1.1.2 Limit of a function 1.1.3 Limit laws 1.1.4 Mathematical definition of a it 1.1.5 Infinite it 1.1. Continuity
More informationf(x 0 + h) f(x 0 ) h slope of secant line = m sec
Derivatives Using limits, we can define the slope of a tangent line to a function. When given a function f(x), and given a point P (x 0, f(x 0 )) on f, if we want to find the slope of the tangent line
More informationDRAFT - Math 101 Lecture Note - Dr. Said Algarni
3 Differentiation Rules 3.1 The Derivative of Polynomial and Exponential Functions In this section we learn how to differentiate constant functions, power functions, polynomials, and exponential functions.
More information2.2 The derivative as a Function
2.2 The derivative as a Function Recall: The derivative of a function f at a fixed number a: f a f a+h f(a) = lim h 0 h Definition (Derivative of f) For any number x, the derivative of f is f x f x+h f(x)
More information2.1 The derivative. Rates of change. m sec = y f (a + h) f (a)
2.1 The derivative Rates of change 1 The slope of a secant line is m sec = y f (b) f (a) = x b a and represents the average rate of change over [a, b]. Letting b = a + h, we can express the slope of the
More information1 The Derivative and Differrentiability
1 The Derivative and Differrentiability 1.1 Derivatives and rate of change Exercise 1 Find the equation of the tangent line to f (x) = x 2 at the point (1, 1). Exercise 2 Suppose that a ball is dropped
More information3.1 Day 1: The Derivative of a Function
A P Calculus 3.1 Day 1: The Derivative of a Function I CAN DEFINE A DERIVATIVE AND UNDERSTAND ITS NOTATION. Last chapter we learned to find the slope of a tangent line to a point on a graph by using a
More informationDRAFT - Math 101 Lecture Note - Dr. Said Algarni
2 Limits 2.1 The Tangent Problems The word tangent is derived from the Latin word tangens, which means touching. A tangent line to a curve is a line that touches the curve and a secant line is a line that
More informationAdvanced Mathematics Unit 2 Limits and Continuity
Advanced Mathematics 3208 Unit 2 Limits and Continuity NEED TO KNOW Expanding Expanding Expand the following: A) (a + b) 2 B) (a + b) 3 C) (a + b)4 Pascals Triangle: D) (x + 2) 4 E) (2x -3) 5 Random Factoring
More informationAdvanced Mathematics Unit 2 Limits and Continuity
Advanced Mathematics 3208 Unit 2 Limits and Continuity NEED TO KNOW Expanding Expanding Expand the following: A) (a + b) 2 B) (a + b) 3 C) (a + b)4 Pascals Triangle: D) (x + 2) 4 E) (2x -3) 5 Random Factoring
More informationAP Calculus ---Notecards 1 20
AP Calculus ---Notecards 1 20 NC 1 For a it to exist, the left-handed it must equal the right sided it x c f(x) = f(x) = L + x c A function can have a it at x = c even if there is a hole in the graph at
More informationSection 1.4 Tangents and Velocity
Math 132 Tangents and Velocity Section 1.4 Section 1.4 Tangents and Velocity Tangent Lines A tangent line to a curve is a line that just touches the curve. In terms of a circle, the definition is very
More informationChapter 3: Derivatives
Name: Date: Period: AP Calc AB Mr. Mellina Chapter 3: Derivatives Sections: v 2.4 Rates of Change & Tangent Lines v 3.1 Derivative of a Function v 3.2 Differentiability v 3.3 Rules for Differentiation
More informationMATH 1902: Mathematics for the Physical Sciences I
MATH 1902: Mathematics for the Physical Sciences I Dr Dana Mackey School of Mathematical Sciences Room A305 A Email: Dana.Mackey@dit.ie Dana Mackey (DIT) MATH 1902 1 / 46 Module content/assessment Functions
More informationChapter 12: Differentiation. SSMth2: Basic Calculus Science and Technology, Engineering and Mathematics (STEM) Strands Mr. Migo M.
Chapter 12: Differentiation SSMth2: Basic Calculus Science and Technology, Engineering and Mathematics (STEM) Strands Mr. Migo M. Mendoza Chapter 12: Differentiation Lecture 12.1: The Derivative Lecture
More informationChapter 4 Notes, Calculus I with Precalculus 3e Larson/Edwards
4.1 The Derivative Recall: For the slope of a line we need two points (x 1,y 1 ) and (x 2,y 2 ). Then the slope is given by the formula: m = y x = y 2 y 1 x 2 x 1 On a curve we can find the slope of a
More informationMATH 2554 (Calculus I)
MATH 2554 (Calculus I) Dr. Ashley K. University of Arkansas February 21, 2015 Table of Contents Week 6 1 Week 6: 16-20 February 3.5 Derivatives as Rates of Change 3.6 The Chain Rule 3.7 Implicit Differentiation
More informationSection 1.1: A Preview of Calculus When you finish your homework, you should be able to
Section 1.1: A Preview of Calculus When you finish your homework, you should be able to π Understand what calculus is and how it compares with precalculus π Understand that the tangent line problem is
More informationBlue Pelican Calculus First Semester
Blue Pelican Calculus First Semester Student Version 1.01 Copyright 2011-2013 by Charles E. Cook; Refugio, Tx Edited by Jacob Cobb (All rights reserved) Calculus AP Syllabus (First Semester) Unit 1: Function
More informationDIFFERENTIATION RULES
3 DIFFERENTIATION RULES DIFFERENTIATION RULES 3. The Product and Quotient Rules In this section, we will learn about: Formulas that enable us to differentiate new functions formed from old functions by
More informationChapter 2: Differentiation
Chapter 2: Differentiation Spring 2018 Department of Mathematics Hong Kong Baptist University 1 / 82 2.1 Tangent Lines and Their Slopes This section deals with the problem of finding a straight line L
More informationLIMITS AND DERIVATIVES
2 LIMITS AND DERIVATIVES LIMITS AND DERIVATIVES 1. Equation In Section 2.7, we considered the derivative of a function f at a fixed number a: f '( a) lim h 0 f ( a h) f ( a) h In this section, we change
More informationMAT137 Calculus! Lecture 6
MAT137 Calculus! Lecture 6 Today: 3.2 Differentiation Rules; 3.3 Derivatives of higher order. 3.4 Related rates 3.5 Chain Rule 3.6 Derivative of Trig. Functions Next: 3.7 Implicit Differentiation 4.10
More informationAverage rates of change May be used to estimate the derivative at a point
Derivatives Big Ideas Rule of Four: Numerically, Graphically, Analytically, and Verbally Average rate of Change: Difference Quotient: y x f( a+ h) f( a) f( a) f( a h) f( a+ h) f( a h) h h h Average rates
More informationSlide 1. Slide 2. Slide 3 Remark is a new function derived from called derivative. 2.2 The derivative as a Function
Slide 1 2.2 The derivative as a Function Slide 2 Recall: The derivative of a function number : at a fixed Definition (Derivative of ) For any number, the derivative of is Slide 3 Remark is a new function
More informationOne-Variable Calculus
POLI 270 - Mathematical and Statistical Foundations Department of Political Science University California, San Diego September 30, 2010 1 s,, 2 al Relationships Political Science, economics, sociology,
More informationChapter 2 Overview: Introduction to Limits and Derivatives
Chapter 2 Overview: Introduction to Limits and Derivatives In a later chapter, maximum and minimum points of a curve will be found both by calculator and algebraically. While the algebra of this process
More informationSection 2.1, Section 3.1 Rate of change, Tangents and Derivatives at a point
Section 2.1, Section 3.1 Rate of change, Tangents and Derivatives at a point Line through P and Q approaches to the tangent line at P as Q approaches P. That is as a + h a = h gets smaller. Slope of the
More informationMATH 116, LECTURE 13, 14 & 15: Derivatives
MATH 116, LECTURE 13, 14 & 15: Derivatives 1 Formal Definition of the Derivative We have seen plenty of limits so far, but very few applications. In particular, we have seen very few functions for which
More informationChapter 1 Functions and Limits
Contents Chapter 1 Functions and Limits Motivation to Chapter 1 2 4 Tangent and Velocity Problems 3 4.1 VIDEO - Secant Lines, Average Rate of Change, and Applications......................... 3 4.2 VIDEO
More information2.1 The Tangent and Velocity Problems
2.1 The Tangent and Velocity Problems Tangents What is a tangent? Tangent lines and Secant lines Estimating slopes from discrete data: Example: 1. A tank holds 1000 gallons of water, which drains from
More informationLearning Objectives for Math 165
Learning Objectives for Math 165 Chapter 2 Limits Section 2.1: Average Rate of Change. State the definition of average rate of change Describe what the rate of change does and does not tell us in a given
More informationChapter 5: Limits and Derivatives
Chapter 5: Limits and Derivatives Chapter 5 Overview: Introduction to Limits and Derivatives In a later chapter, maximum and minimum points of a curve will be found both by calculator and algebraically.
More informationTangent Lines Sec. 2.1, 2.7, & 2.8 (continued)
Tangent Lines Sec. 2.1, 2.7, & 2.8 (continued) Prove this Result How Can a Derivative Not Exist? Remember that the derivative at a point (or slope of a tangent line) is a LIMIT, so it doesn t exist whenever
More informationUnit 1 PreCalculus Review & Limits
1 Unit 1 PreCalculus Review & Limits Factoring: Remove common factors first Terms - Difference of Squares a b a b a b - Sum of Cubes ( )( ) a b a b a ab b 3 3 - Difference of Cubes a b a b a ab b 3 3 3
More information( ) as a fraction. If both numerator and denominator are
A. Limits and Horizontal Asymptotes What you are finding: You can be asked to find lim f x x a (H.A.) problem is asking you find lim f x x ( ) and lim f x x ( ). ( ) or lim f x x ± ( ). Typically, a horizontal
More informationDerivatives and Rates of Change
Sec.1 Derivatives and Rates of Change A. Slope of Secant Functions rise Recall: Slope = m = = run Slope of the Secant Line to a Function: Examples: y y = y1. From this we are able to derive: x x x1 m y
More informationMath 134 Exam 2 November 5, 2009
Math 134 Exam 2 November 5, 2009 Name: Score: / 80 = % 1. (24 Points) (a) (8 Points) Find the slope of the tangent line to the curve y = 9 x2 5 x 2 at the point when x = 2. To compute this derivative we
More information1.1 Radical Expressions: Rationalizing Denominators
1.1 Radical Expressions: Rationalizing Denominators Recall: 1. A rational number is one that can be expressed in the form a, where b 0. b 2. An equivalent fraction is determined by multiplying or dividing
More informationChapter 2: Differentiation
Chapter 2: Differentiation Winter 2016 Department of Mathematics Hong Kong Baptist University 1 / 75 2.1 Tangent Lines and Their Slopes This section deals with the problem of finding a straight line L
More information2.1 How Do We Measure Speed? Student Notes HH6ed. Time (sec) Position (m)
2.1 How Do We Measure Speed? Student Notes HH6ed Part I: Using a table of values for a position function The table below represents the position of an object as a function of time. Use the table to answer
More informationSec 4.1 Limits, Informally. When we calculated f (x), we first started with the difference quotient. f(x + h) f(x) h
1 Sec 4.1 Limits, Informally When we calculated f (x), we first started with the difference quotient f(x + h) f(x) h and made h small. In other words, f (x) is the number f(x+h) f(x) approaches as h gets
More informationDIFFERENTIATION RULES
3 DIFFERENTIATION RULES DIFFERENTIATION RULES We have: Seen how to interpret derivatives as slopes and rates of change Seen how to estimate derivatives of functions given by tables of values Learned how
More informationTangent Lines and Derivatives
The Derivative and the Slope of a Graph Tangent Lines and Derivatives Recall that the slope of a line is sometimes referred to as a rate of change. In particular, we are referencing the rate at which the
More informationAPPLICATIONS OF DIFFERENTIATION
4 APPLICATIONS OF DIFFERENTIATION APPLICATIONS OF DIFFERENTIATION 4.9 Antiderivatives In this section, we will learn about: Antiderivatives and how they are useful in solving certain scientific problems.
More informationLimit. Chapter Introduction
Chapter 9 Limit Limit is the foundation of calculus that it is so useful to understand more complicating chapters of calculus. Besides, Mathematics has black hole scenarios (dividing by zero, going to
More informationAP Calculus. Derivatives.
1 AP Calculus Derivatives 2015 11 03 www.njctl.org 2 Table of Contents Rate of Change Slope of a Curve (Instantaneous ROC) Derivative Rules: Power, Constant, Sum/Difference Higher Order Derivatives Derivatives
More informationFree Response Questions Compiled by Kaye Autrey for face-to-face student instruction in the AP Calculus classroom
Free Response Questions 1969-010 Compiled by Kaye Autrey for face-to-face student instruction in the AP Calculus classroom 1 AP Calculus Free-Response Questions 1969 AB 1 Consider the following functions
More informationSolution: It could be discontinuous, or have a vertical tangent like y = x 1/3, or have a corner like y = x.
1. Name three different reasons that a function can fail to be differentiable at a point. Give an example for each reason, and explain why your examples are valid. It could be discontinuous, or have a
More informationChapter 4. Section Derivatives of Exponential and Logarithmic Functions
Chapter 4 Section 4.2 - Derivatives of Exponential and Logarithmic Functions Objectives: The student will be able to calculate the derivative of e x and of lnx. The student will be able to compute the
More informationAP Calculus Worksheet: Chapter 2 Review Part I
AP Calculus Worksheet: Chapter 2 Review Part I 1. Given y = f(x), what is the average rate of change of f on the interval [a, b]? What is the graphical interpretation of your answer? 2. The derivative
More informationAP Calculus Free-Response Questions 1969-present AB
AP Calculus Free-Response Questions 1969-present AB 1969 1. Consider the following functions defined for all x: f 1 (x) = x, f (x) = xcos x, f 3 (x) = 3e x, f 4 (x) = x - x. Answer the following questions
More information1 Antiderivatives graphically and numerically
Math B - Calculus by Hughes-Hallett, et al. Chapter 6 - Constructing antiderivatives Prepared by Jason Gaddis Antiderivatives graphically and numerically Definition.. The antiderivative of a function f
More informationMATH 114 Calculus Notes on Chapter 2 (Limits) (pages 60-? in Stewart)
Still under construction. MATH 114 Calculus Notes on Chapter 2 (Limits) (pages 60-? in Stewart) As seen in A Preview of Calculus, the concept of it underlies the various branches of calculus. Hence we
More informationDifferentiation - Quick Review From Calculus
Differentiation - Quick Review From Calculus Philippe B. Laval KSU Current Semester Philippe B. Laval (KSU) Differentiation - Quick Review From Calculus Current Semester 1 / 13 Introduction In this section,
More informationTopics and Concepts. 1. Limits
Topics and Concepts 1. Limits (a) Evaluating its (Know: it exists if and only if the it from the left is the same as the it from the right) (b) Infinite its (give rise to vertical asymptotes) (c) Limits
More informationOBJECTIVE Find limits of functions, if they exist, using numerical or graphical methods.
1.1 Limits: A Numerical and Graphical Approach OBJECTIVE Find limits of functions, if they exist, using numerical or graphical methods. 1.1 Limits: A Numerical and Graphical Approach DEFINITION: As x approaches
More informationter. on Can we get a still better result? Yes, by making the rectangles still smaller. As we make the rectangles smaller and smaller, the
Area and Tangent Problem Calculus is motivated by two main problems. The first is the area problem. It is a well known result that the area of a rectangle with length l and width w is given by A = wl.
More informationChapter 5 - Differentiating Functions
Chapter 5 - Differentiating Functions Section 5.1 - Differentiating Functions Differentiation is the process of finding the rate of change of a function. We have proven that if f is a variable dependent
More information10550 PRACTICE FINAL EXAM SOLUTIONS. x 2 4. x 2 x 2 5x +6 = lim x +2. x 2 x 3 = 4 1 = 4.
55 PRACTICE FINAL EXAM SOLUTIONS. First notice that x 2 4 x 2x + 2 x 2 5x +6 x 2x. This function is undefined at x 2. Since, in the it as x 2, we only care about what happens near x 2 an for x less than
More information1 + x 2 d dx (sec 1 x) =
Page This exam has: 8 multiple choice questions worth 4 points each. hand graded questions worth 4 points each. Important: No graphing calculators! Any non-graphing, non-differentiating, non-integrating
More informationCalculus I Review Solutions
Calculus I Review Solutions. Compare and contrast the three Value Theorems of the course. When you would typically use each. The three value theorems are the Intermediate, Mean and Extreme value theorems.
More informationChapter 2 Derivatives
Contents Chapter 2 Derivatives Motivation to Chapter 2 2 1 Derivatives and Rates of Change 3 1.1 VIDEO - Definitions................................................... 3 1.2 VIDEO - Examples and Applications
More informationAnnouncements. Topics: Homework:
Topics: Announcements - section 2.6 (limits at infinity [skip Precise Definitions (middle of pg. 134 end of section)]) - sections 2.1 and 2.7 (rates of change, the derivative) - section 2.8 (the derivative
More informationMATH 100 and MATH 180 Learning Objectives Session 2010W Term 1 (Sep Dec 2010)
Course Prerequisites MATH 100 and MATH 180 Learning Objectives Session 2010W Term 1 (Sep Dec 2010) As a prerequisite to this course, students are required to have a reasonable mastery of precalculus mathematics
More informationReview for the Final Exam
Math 171 Review for the Final Exam 1 Find the limits (4 points each) (a) lim 4x 2 3; x x (b) lim ( x 2 x x 1 )x ; (c) lim( 1 1 ); x 1 ln x x 1 sin (x 2) (d) lim x 2 x 2 4 Solutions (a) The limit lim 4x
More informationWORKBOOK. MATH 31. CALCULUS AND ANALYTIC GEOMETRY I.
WORKBOOK. MATH 31. CALCULUS AND ANALYTIC GEOMETRY I. DEPARTMENT OF MATHEMATICS AND COMPUTER SCIENCE Contributors: U. N. Iyer and P. Laul. (Many problems have been directly taken from Single Variable Calculus,
More informationVIDEO LINKS: a) b)
CALCULUS 30: OUTCOME 4A DAY 1 SLOPE AND RATE OF CHANGE To review the concepts of slope and rate of change. VIDEO LINKS: a) https://goo.gl/r9fhx3 b) SLOPE OF A LINE: Is a measure of the steepness of a line
More informationCHAPTER 3 DIFFERENTIATION
CHAPTER 3 DIFFERENTIATION 3.1 THE DERIVATIVE AND THE TANGENT LINE PROBLEM You will be able to: - Find the slope of the tangent line to a curve at a point - Use the limit definition to find the derivative
More informationMath Review ECON 300: Spring 2014 Benjamin A. Jones MATH/CALCULUS REVIEW
MATH/CALCULUS REVIEW SLOPE, INTERCEPT, and GRAPHS REVIEW (adapted from Paul s Online Math Notes) Let s start with some basic review material to make sure everybody is on the same page. The slope of a line
More informationChapter 3 Derivatives
Chapter Derivatives Section 1 Derivative of a Function What you ll learn about The meaning of differentiable Different ways of denoting the derivative of a function Graphing y = f (x) given the graph of
More informationGoal: Approximate the area under a curve using the Rectangular Approximation Method (RAM) RECTANGULAR APPROXIMATION METHODS
AP Calculus 5. Areas and Distances Goal: Approximate the area under a curve using the Rectangular Approximation Method (RAM) Exercise : Calculate the area between the x-axis and the graph of y = 3 2x.
More informationFor those of you who are taking Calculus AB concurrently with AP Physics, I have developed a
AP Physics C: Mechanics Greetings, For those of you who are taking Calculus AB concurrently with AP Physics, I have developed a brief introduction to Calculus that gives you an operational knowledge of
More informationa x a y = a x+y a x a = y ax y (a x ) r = a rx and log a (xy) = log a (x) + log a (y) log a ( x y ) = log a(x) log a (y) log a (x r ) = r log a (x).
You should prepare the following topics for our final exam. () Pre-calculus. (2) Inverses. (3) Algebra of Limits. (4) Derivative Formulas and Rules. (5) Graphing Techniques. (6) Optimization (Maxima and
More informationMath 147 Exam II Practice Problems
Math 147 Exam II Practice Problems This review should not be used as your sole source for preparation for the exam. You should also re-work all examples given in lecture, all homework problems, all lab
More informationFormulas that must be memorized:
Formulas that must be memorized: Position, Velocity, Acceleration Speed is increasing when v(t) and a(t) have the same signs. Speed is decreasing when v(t) and a(t) have different signs. Section I: Limits
More informationCalculus AB Topics Limits Continuity, Asymptotes
Calculus AB Topics Limits Continuity, Asymptotes Consider f x 2x 1 x 3 1 x 3 x 3 Is there a vertical asymptote at x = 3? Do not give a Precalculus answer on a Calculus exam. Consider f x 2x 1 x 3 1 x 3
More informationm(x) = f(x) + g(x) m (x) = f (x) + g (x) (The Sum Rule) n(x) = f(x) g(x) n (x) = f (x) g (x) (The Difference Rule)
Chapter 3 Differentiation Rules 3.1 Derivatives of Polynomials and Exponential Functions Aka The Short Cuts! Yay! f(x) = c f (x) = 0 g(x) = x g (x) = 1 h(x) = x n h (x) = n x n-1 (The Power Rule) k(x)
More informationMA 137 Calculus 1 with Life Science Applications The Chain Rule and Higher Derivatives (Section 4.4)
MA 137 Calculus 1 with Life Science Applications and (Section 4.4) Alberto Corso alberto.corso@uky.edu Department of Mathematics University of Kentucky March 2, 2016 1/15 Theorem Rules of Differentiation
More information2.1 Derivatives and Rates of Change
1a 1b 2.1 Derivatives an Rates of Change Tangent Lines Example. Consier y f x x 2 0 2 x-, 0 4 y-, f(x) axes, curve C Consier a smooth curve C. A line tangent to C at a point P both intersects C at P an
More informationAP Calculus AB: Semester Review Notes Information in the box are MASTERY CONCEPTS. Be prepared to apply these concepts on your midterm.
AP Calculus AB: Semester Review Notes Information in the box are MASTERY CONCEPTS. Be prepared to apply these concepts on your midterm. Name: Date: Period: I. Limits and Continuity Definition of Average
More informationChapter 2 Differentiation. 2.1 Tangent Lines and Their Slopes. Calculus: A Complete Course, 8e Chapter 2: Differentiation
Chapter 2 Differentiation 2.1 Tangent Lines and Their Slopes 1) Find the slope of the tangent line to the curve y = 4x x 2 at the point (-1, 0). A) -1 2 C) 6 D) 2 1 E) -2 2) Find the equation of the tangent
More informationMATH 408N PRACTICE FINAL
2/03/20 Bormashenko MATH 408N PRACTICE FINAL Show your work for all the problems. Good luck! () Let f(x) = ex e x. (a) [5 pts] State the domain and range of f(x). Name: TA session: Since e x is defined
More information3.4 The Chain Rule. F (x) = f (g(x))g (x) Alternate way of thinking about it: If y = f(u) and u = g(x) where both are differentiable functions, then
3.4 The Chain Rule To find the derivative of a function that is the composition of two functions for which we already know the derivatives, we can use the Chain Rule. The Chain Rule: Suppose F (x) = f(g(x)).
More informationMATH 1241 Common Final Exam Fall 2010
MATH 1241 Common Final Exam Fall 2010 Please print the following information: Name: Instructor: Student ID: Section/Time: The MATH 1241 Final Exam consists of three parts. You have three hours for the
More informationFoundations of Calculus. November 18, 2014
Foundations of Calculus November 18, 2014 Contents 1 Conic Sections 3 11 A review of the coordinate system 3 12 Conic Sections 4 121 Circle 4 122 Parabola 5 123 Ellipse 5 124 Hyperbola 6 2 Review of Functions
More informationTo take the derivative of x raised to a power, you multiply in front by the exponent and subtract 1 from the exponent.
MA123, Chapter 5: Formulas for derivatives (pp. 83-102) Date: Chapter Goals: Know and be able to apply the formulas for derivatives. Understand the chain rule and be able to apply it. Know how to compute
More informationAP Calculus Chapter 3 Testbank (Mr. Surowski)
AP Calculus Chapter 3 Testbank (Mr. Surowski) Part I. Multiple-Choice Questions (5 points each; please circle the correct answer.). If f(x) = 0x 4 3 + x, then f (8) = (A) (B) 4 3 (C) 83 3 (D) 2 3 (E) 2
More informationLIMITS AND DERIVATIVES
2 LIMITS AND DERIVATIVES LIMITS AND DERIVATIVES The idea of a limit underlies the various branches of calculus. It is therefore appropriate to begin our study of calculus by investigating limits and their
More informationTable of Contents. Module 1
Table of Contents Module Order of operations 6 Signed Numbers Factorization of Integers 7 Further Signed Numbers 3 Fractions 8 Power Laws 4 Fractions and Decimals 9 Introduction to Algebra 5 Percentages
More informationAnna D Aloise May 2, 2017 INTD 302: Final Project. Demonstrate an Understanding of the Fundamental Concepts of Calculus
Anna D Aloise May 2, 2017 INTD 302: Final Project Demonstrate an Understanding of the Fundamental Concepts of Calculus Analyzing the concept of limit numerically, algebraically, graphically, and in writing.
More informationChapter 2 THE DERIVATIVE
Chapter 2 THE DERIVATIVE 2.1 Two Problems with One Theme Tangent Line (Euclid) A tangent is a line touching a curve at just one point. - Euclid (323 285 BC) Tangent Line (Archimedes) A tangent to a curve
More informationMA4001 Engineering Mathematics 1 Lecture 15 Mean Value Theorem Increasing and Decreasing Functions Higher Order Derivatives Implicit Differentiation
MA4001 Engineering Mathematics 1 Lecture 15 Mean Value Theorem Increasing and Decreasing Functions Higher Order Derivatives Implicit Differentiation Dr. Sarah Mitchell Autumn 2014 Rolle s Theorem Theorem
More informationSpring 2015 Sample Final Exam
Math 1151 Spring 2015 Sample Final Exam Final Exam on 4/30/14 Name (Print): Time Limit on Final: 105 Minutes Go on carmen.osu.edu to see where your final exam will be. NOTE: This exam is much longer than
More information3 Geometrical Use of The Rate of Change
Arkansas Tech University MATH 224: Business Calculus Dr. Marcel B. Finan Geometrical Use of The Rate of Change Functions given by tables of values have their limitations in that nearly always leave gaps.
More informationPreliminaries Lectures. Dr. Abdulla Eid. Department of Mathematics MATHS 101: Calculus I
Preliminaries 2 1 2 Lectures Department of Mathematics http://www.abdullaeid.net/maths101 MATHS 101: Calculus I (University of Bahrain) Prelim 1 / 35 Pre Calculus MATHS 101: Calculus MATHS 101 is all about
More informationDifferential Calculus Average Rate of Change (AROC) The average rate of change of y over an interval is equal to
Differential Calculus Average Rate of Change (AROC) The average rate of change of y over an interval is equal to change in y y y1 f ( x) f ( x1 ) f ( b) f ( a). changein x x x x x b a 1 Example: Find the
More informationfunction independent dependent domain range graph of the function The Vertical Line Test
Functions A quantity y is a function of another quantity x if there is some rule (an algebraic equation, a graph, a table, or as an English description) by which a unique value is assigned to y by a corresponding
More informationDecember Exam Summary
December Exam Summary 1 Lines and Distances 1.1 List of Concepts Distance between two numbers on the real number line or on the Cartesian Plane. Increments. If A = (a 1, a 2 ) and B = (b 1, b 2 ), then
More information