DRAFT CHAPTER : Derivatives Errors will be corrected before printing. Final book will be available August 008.
Chapter DERIVATIVES Imagine a driver speeding down a highway, at 140 km/h. He hears a police siren and is quickly pulled over. The police officer tells him that he was speeding, but the driver argues that because he has travelled 00 km from home in two hours, his average speed is within the 100 km/h limit. The driver s argument fails because police officers charge speeders based on their instantaneous speed, not their average speed. There are many other situations in which the instantaneous rate of change is more important than the average rate of change. In calculus, the derivative is a tool for finding instantaneous rates of change. This chapter shows how the derivative can be determined and applied in a great variety of situations. CHAPTER EXPECTATIONS In this chapter, you will understand and determine derivatives of polynomial and simple rational functions from first principles, Section.1 identify eamples of functions that are not differentiable, Section.1 justify and use the rules for determining derivatives, Sections.,.3,.4,.5 identify composition as two functions applied in succession, Section.5 determine the composition of two functions epressed in notation, and decompose a given composite function into its parts, Section.5 use the derivative to solve problems involving instantaneous rates of change Sections.,.3,.4,.5
Review of Prerequisite Skills Before beginning your study of derivatives, it may be helpful to review the following concepts from previous courses and the previous chapter: Working with the properties of eponents Simplifying radical epressions Finding the slopes of parallel and perpendicular lines Simplifying rational epressions Epanding and factoring algebraic epressions Evaluating epressions Working with the difference quotient Eercise 3 y 5 1. Use the eponent laws to simplify each of the following epressions. Epress your answers with positive eponents. a. a 5 a 3 c. 4p 7 6p 9 1p 15 e. 13e 6 1e 3 4 b. 1 a 3 d. 1a 4 b 5 1a 6 b f. 13a 4 3a 3 1 b 3 4 1a 5 b. Simplify and write each epression in eponential form. a. 1 1 1 3 b. 18 6 3 c. Va Va 3 Va 3. Determine the slope of a line that is perpendicular to a line with each given slope. a. 5 b. 1 c. 3 3 d. 1 4. Determine the equation of each of the following lines: a. passing through points A1 3, 4 and B19, b. passing through point A1, 5 and parallel to the line c. perpendicular to the line with y 3 and passing 4 6 through point A14, 3 REVIEW OF PREREQUISITE SKILLS
5. Epand, and collect like terms. a. 1 3y1 y d. 1 y 513 8y b. 1 1 3 4 e. 1 3y 15 y c. 16 31 7 f. 31 y 15 y15 y 6. Simplify each epression. a. 31 5 3 1 y1 y 1 y3 d. 1 51 y 10 b. y 1y 5 7 e. 5 1y 1y 5 4y 3 1 c. 4 1 f. 1h k 9 1h k 3 7. Factor each epression completely. a. 4k 9 c. 3a 4a 7 e. 3 y 3 b. 4 3 d. 4 1 f. r 4 5r 4 8. Use the factor theorem to factor the following epressions: a. a 3 b 3 b. a 5 b 5 c. a 7 b 7 d. a n b n 9. If f 1 4 3 7, evaluate a. f 1 b. f 1 1 c. f a 1 d. b f 1 0.5 10. Rationalize the denominator in each of the following epressions: 3 4 V 3V 3V 4V3 a. b. c. d. V V3 3 4V 3V 4V3 11. a. If f 1 3 determine the epression for the difference quotient f 1a h f 1a when a. Eplain what this epression can be used for. h b. Evaluate the epression you found in part a for a small value of h such as h 0.01. c. Eplain what the value you determined in part b represents. CHAPTER 3
CAREER LINK Investigate CHAPTER : THE ELASTICITY OF DEMAND Have you ever wondered how businesses set prices for their goods and services? An important idea in marketing is elasticity of demand, or the response of consumers to a change in price. Consumers respond differently to a change in the price of a staple item, such as bread, than they do to a change in the price of a luury item, such as jewellery. A family would probably still buy the same quantity of bread if the price increased by 0%. This is called inelastic demand. If the price of a gold chain, however, increased by 0%, sales would likely decrease 40% or more. This is called elastic demand. Mathematically, elasticity is defined as the negative of the relative (percentage) change in the number demanded Q n n R divided by the relative (percentage) change in the price For eample, if a store increased the price of a CD from $17.99 to $19.99, and the number sold per week went from 10 to 80, the elasticity would be E ca An elasticity of about 3 means that the change in demand is three times as large, in percentage terms, as the change in price. The CDs have an elastic demand because a small change in price can cause a large change in demand. In general, goods or services with elasticities greater than one 1E 7 1 are considered elastic (e.g., new cars), and those with elasticities less than one 1E 6 1 are considered inelastic (e.g., milk). In our eample, we calculated the average elasticity between two price levels, but, in reality, businesses want to know the elasticity at a specific, or instantaneous, price level. In this chapter, you will develop the rules of differentiation that will enable you to calculate the instantaneous rate of change for several classes of functions. Case Study Marketer: Product Pricing In addition to developing advertising strategies, marketing departments also conduct research into, and make decisions on, pricing. Suppose that the demand price relationship for weekly movie rentals at a convenience store is n1p 500 where n1p is demand and p is price. p, DISCUSSION QUESTIONS E ca n p b a n p bd 80 10 19.99 17.99 b a bd 3.00 10 17.99 Q p p R: 1. Generate two lists, each with at least five goods and services that you have purchased recently, classifying each of the goods and services as having elastic or inelastic demand.. Calculate and discuss the elasticity if a movie rental fee increases from $1.99 to $.99. 4 CAREER LINK
Section.1 The Derivative Function In this chapter, we will etend the concepts of the slope of a tangent and the rate of change to introduce the derivative. We will eamine the methods of differentiation, which we can use to determine the derivatives of polynomial and rational functions. These methods include the use of the power rule, sum and difference rules, and product and quotient rules, as well as the chain rule for the composition of functions. The Derivative at a Point f 1a h f 1a In the previous chapter, we encountered limits of the form lim. hs0 h This limit has two interpretations: the slope of the tangent to the graph y f 1 at the point 1a, f 1a, and the instantaneous rate of change of y f 1 with respect to at a. Since this limit plays a central role in calculus, it is given a name and a concise notation. It is called the derivative of f() at a. It is denoted by f 1a and is read as f prime of a. The derivative of f at the number a is given by provided that this limit eists. f 1a lim hs0 f 1a h f 1a, h EXAMPLE 1 Selecting a limit strategy to determine the derivative at a number Determine the derivative of f 1 at 3. Solution Using the definition, the derivative at f 1 3 h f 1 3 f 1 3 lim 1 3 h 1 3 lim 9 6h h 9 lim h1 6 h lim lim 1 6 h hs0 6 3is given by (Epand) (Simplify and Factor) Therefore, the derivative of f 1 at 3is 6. CHAPTER 5
An alternative way of writing the derivative of f at the number a is f 1 f 1a f 1a lim. Sa a In applications where we are required to find the value of the derivative for a number of particular values of, using the definition repeatedly for each value is tedious. The net eample illustrates the efficiency of calculating the derivative of f 1 at an arbitrary value of and using the result to determine the derivatives at a number of particular -values. EXAMPLE Connecting the derivative of a function to an arbitrary value a. Determine the derivative of f 1 at an arbitrary value of. b. Determine the slopes of the tangents to the parabola y at, 0, and 1. Solution a. Using the definition, f 1 h f 1 f 1 lim 1 h lim h h lim h1 h lim lim 1 h hs0 (Epand) (Simplify and Factor) The derivative of f 1 at an arbitrary value of is f 1. b. The required slopes of the tangents to y are obtained by evaluating the derivative f 1 at the given -values. We obtain the slopes by substituting for : f 1 4 f 10 0 f 11 The slopes are 4, 0, and, respectively. 6.1 THE DERIVATIVE FUNCTION In fact, knowing the -coordinate of a point on the parabola y, we can easily find the slope of the tangent at that point. For eample, given the -coordinates of points on the curve, we can produce the following table.
For the Parabola f1 The slope of the tangent to the curve f 1 P(, y) -Coordinate of P Slope of Tangent at P at a point P1, y is given by the derivative f 1. For each -value, there is an 1, 4 1 4 associated value. 1 1, 1 1 10, 0 0 0 11, 1 1 1, 4 4 1a, a a a The graphs of f 1 and the derivative function f 1 are shown below. The tangents at, 0, and 1 are shown on the graph of f 1. 4 4 0 4 y f() = 4 3 1 0 1 ( 1, ) 3 Notice that the graph of the derivative function of the quadratic function (of degree two) is a linear function (of degree one). 3 1 y 1 (1, ) f 9() = 3 INVESTIGATION A. Determine the derivative with respect to of each of the following functions: 5 a. f 1 3 b. f 1 4 c. f 1 B. In Eample, we showed that the derivative of f 1 is f 1. Referring to step 1, what pattern do you see developing? C. Use the pattern from step to predict the derivative of f 1 39. D.What do you think f 1 would be for f 1 n, where n is a positive integer? The Derivative Function The derivative of f at a is a number f 1a. If we let a be arbitrary and assume a general value in the domain of f, the derivative f is a function. For eample, if f 1, f 1, which is itself a function. CHAPTER 7
The Definition of the Derivative Function The derivative of f 1 with respect to is the function f 1, where f 1 h f 1 f 1 lim, provided that this limit eists. hs0 h The limit f 1 is a notation that was developed by Joseph Louis Lagrange (1736 1813), a French mathematician. When you use this limit to determine the derivative of a function, it is called determining the derivative from first principles. In Chapter 1, we discussed velocity at a point. We can now define (instantaneous) velocity as the derivative of position with respect to time. If the position of a body at time t is s1t, then the velocity of the body at time t is v1t s 1t lim hs0 s1t h s 1t. h Likewise, the (instantaneous) rate of change of f 1 with respect to is the f 1 h f 1 function f 1, whose value is f 1 lim. hs0 h EXAMPLE 3 Determining the derivative from first principles Determine the derivative f 1t of the function f 1t Vt, t 0. Solution Using the definition, f 1t h f 1t f 1t lim Vt h Vt lim Vt h Vt lim 1t h t lim 1 Vt h Vt h lim 1 Vt h Vt lim hs0 1 Vt h Vt 1 Vt, for t 7 0 a Vt h Vt Vt h Vt b (Rationalize the numerator) 8.1 THE DERIVATIVE FUNCTION
Note that f 1t Vt is defined for all instances of t 0, whereas its derivative f 1t 1 is defined only for instances when t 7 0. From this, we can see that Vt a function need not have a derivative throughout its entire domain. EXAMPLE 4 Selecting a strategy involving the derivative to determine the equation of a tangent Determine an equation of the tangent to the graph of f 1 1 at the point where. Solution When, y 1 The graph of y 1 y the., 4 y = 1 point Q, 1, and the tangent at the point are shown. R tangent, 1 First find f 1. 4 0 f 1 h f 1 f 1 lim 1 h 1 lim lim hs0 1 h 1 h lim 1 h 1 lim hs0 1 h h h 1 h (Simplify the fraction) 4 4 1 The slope of the tangent at 1 is m f 1 The equation of the tangent is y 1 4. or, in standard form, 4y 4 0. 1 1 4 CHAPTER 9
EXAMPLE 5 Selecting a strategy involving the derivative to solve a problem Determine an equation of the line that is perpendicular to the tangent to the graph of f 1 1 at and that intersects it at the point of tangency. Solution In Eample 4, we found that the slope of the tangent at = is f 1 1 and 4, 1 the point of tangency is Q, The perpendicular line has slope 4, the negative R. reciprocal of 1 Therefore, the required equation is y 1 41 or 4. 8 y 15 0. tangent 4 4 0 y y = 1 normal, 1 4 4 The line whose equation we found in Eample 5 has a name. The normal to the graph of f at point P is the line that is perpendicular to the tangent at P. The Eistence of Derivatives A function f is said to be differentiable at a if f 1a eists. At points where f is not differentiable, we say that the derivative does not eist. Three common ways for a derivative to fail to eist are shown. y y y a a a 10.1 THE DERIVATIVE FUNCTION Cusp Vertical Tangent Discontinuity
EXAMPLE 6 Reasoning about differentiability at a point Show that the absolute value function f 1 is not differentiable at 0. Solution The graph of f 1 is shown. Because the slope for 6 0 is 1, whereas the slope for 7 0 is 1, the graph has a corner at 10, 0, which prevents a unique tangent from being drawn there. We can show this using the definition of a derivative. 4 y = 4 0 4 4 y f 10 h f 10 f 10 lim Now, we will consider one-sided limits. 0h 0 h when h 7 0 and 0h 0 hwhen h 6 0. 0 h 0 lim hs0 h 0h 0 lim hs0 h f 1h 0 lim 0 h 0 lim h lim hs0 h h lim hs0 h lim 1 1 1 hs0 lim hs0 11 1 Since the left-hand limit and the right-hand limit are not the same, the derivative does not eist at 0. From Eample 6, we conclude that it is possible for a function to be continuous at a point and yet not differentiable at this point. CHAPTER 11
Other Notation for Derivatives Symbols other than f 1 are often used to denote the derivative. If y f 1, the dy dy symbols y and are used instead of f 1. The notation was originally used d d by Leibniz and is read dee y by dee. For eample, if y, the derivative is y dy or, in Leibniz notation,. Similarly, in Eample 4, we showed dy that if y 1 d then. The Leibniz notation reminds us of the process by, d 1 which the derivative is obtained namely, as the limit of a difference quotient: dy y lim d S0. By omitting y and f altogether, we can combine these statements and write d which is read the derivative of with respect to is. It is d 1, dy important to note that is not a fraction. d IN SUMMARY Key Ideas f1a h f1a The derivative of a function f at a point 1a, f1a is f 1a lim, f1 f1a or f 1a lim if the limit eists. Sa a A function is said to be differentiable at a if f 1a eists. A function is differentiable on an interval if it is differentiable at every number in the interval. The derivative function for any function f1 is given by f1 h f1 f 1 lim, if the limit eists. Need to Know f1a h f1a To find the derivative at a point a, you can use lim. The derivative f 1a can be interpreted as either the slope of the tangent at 1a, f1a, or the instantaneous rate of change of f1 with respect to when a. Other notations for the derivative of the function y f1 are f 1, y, and dy d. 1.1 THE DERIVATIVE FUNCTION
Eercise.1 PART A 1. State the domain on which f is differentiable. a. y d. 4 4 0 4 y = f() 4 4 y y = f() 4 0 4 4 b. y e. 4 y = f() 4 0 4 4 y 4 y = f() 4 0 4 4 c. y f. 4 4 0 4 4 y = f() y 4 y = f() 4 0 4 4 C. Eplain what the derivative of a function represents. 3. Illustrate two situations in which a function does not have a derivative at 1. 4. For each function, find f 1a h and f 1a h f 1a. a. f 1 5 d. f 1 6 b. f 1 3 1 e. f 1 7 4 c. f 1 3 4 1 f. f 1 4 CHAPTER 13
K PART B 5. For each function, find the value of the derivative f 1a for the given value of a. a. f 1, a 1 c. f 1 V 1, a 0 b. f 1 3 1, a 3 d. f 1 5, a 1 6. Use the definition of the derivative to find f 1 for each function. a. f 1 5 8 c. f 1 6 3 7 b. f 1 4 d. f 1 V3 dy 7. In each case, find the derivative from first principles. d a. y 6 7 b. y 1 1 c. y 3 8. Determine the slope of the tangents to y 4 when 0, 1, and. Sketch the graph, showing these tangents. A 9. a. Sketch the graph of f 1 3. b. Calculate the slopes of the tangents to f 1 3 at points with -coordinates, 1, 0, 1,. c. Sketch the graph of the derivative function f 1. d. Compare the graphs of f 1 and f 1. 10. An object moves in a straight line with its position at time t seconds given by s1t t 8t, where s is measured in metres. Find the velocity when t 0, t 4, and t 6. 11. Determine an equation of the line that is tangent to the graph of f 1 V 1 and parallel to 6y 4 0. dy 1. For each function, use the definition of the derivative to determine where d, a, b, c, and m are constants. a. y c c. y m b b. y d. y a b c 13. Does the function f 1 3 ever have a negative slope? If so, where? Give reasons for your answer. 14. A football is kicked up into the air. Its height, h, above the ground in metres at t seconds can be modelled by h1t 18t 4.9t. a. Determine h 1. b. What does h 1 represent? 14.1 THE DERIVATIVE FUNCTION
T 15. Match each function in graphs a, b, and c with its corresponding derivative, graphed in d, e, and f. a. y d. y 4 4 0 4 4 4 4 0 4 4 b. y e. 4 4 0 4 4 y 4 4 0 4 4 c. y f. 4 4 0 4 4 y 4 4 0 4 4 PART C 16. For the function f 1 0 0, show that f 10 eists. What is the value? 17. If f 1a 0 and f 1a 6, find lim hs0 f 1a h. h 18. Give an eample of a function that is continuous on q 6 6qbut is not differentiable at 3. 19. At what point on the graph of y 4 5 is the tangent parallel to y 1? 0. Determine the equations of both lines that are tangent to the graph of f 1 and pass through point 11, 3. CHAPTER 15
Section. The Derivatives of Polynomial Functions We have seen that derivatives of functions are of practical use because they represent instantaneous rates of change. Computing derivatives from the limit definition, as we did in Section.1, is tedious and time-consuming. In this section, we will develop some rules that simplify the process of differentiation. We will begin developing the rules of differentiation by looking at the constant function, f 1 k. Since the graph of any constant function is a horizontal line with slope zero at each point, the derivative should be zero. For eample, d if f 1 4, then f 13 0. Alternatively, we can write 13 0. d 4 4 0 4 f() = 4 slope = 0 4 (3, 4) y The Constant Function Rule If f 1 k, where k is a constant, then f 1 0. In Leibniz notation, d d 1k 0. Proof: f 1 lim hs0 f 1 h f 1 h k k lim lim 0 hs0 0 (Since f 1 k and f 1 h k for all h) 16. THE DERIVATIVES OF POLYNOMIAL FUNCTIONS
EXAMPLE 1 Applying the constant function rule a. If f 1 5, f 1 0. b. If y p, dy 0. d A power function is a function of the form f 1 n, where n is a real number. In the previous section, we observed that for f 1, f 1 ; for g 1 V 1 g 1 1 ; and for h1 1 1, h 1. 1 1, V d As well, we hypothesized that In fact, this is true and is called d 1n n n 1. the power rule. The Power Rule If f 1 n, where n is a real number, then f 1 n n 1. In Leibniz notation, d d 1n n n 1. Proof: (Note: n is a positive integer.) Using the definition of the derivative, f 1 lim hs0 f 1 h f 1, where f 1 n h 1 h n n lim 1 h 31 h n 1 1 h n p lim hs0 h 1 h n n 1 4 h lim 31 h n 1 1 h n p 1 h n n 1 4 hs0 n 1 n 1 p 1 n n 1 n 1 n 1 p n 1 n 1 n n 1 (Factor) (Divide out h) (Since there are n terms) CHAPTER 17
EXAMPLE Applying the power rule a. If f 1 7, then f 1 7 6. b. If g1 1 then g 1 3 3 1 3 4 3 3 3, 4. 3 ds c. If s t, dt 3 t 1 3 Vt. d d. d 1 11 1 0 1 The Constant Multiple Rule If f 1 kg1, where k is a constant, then f 1 kg 1. In Leibniz notation, d 1ky kdy d d. Proof: Let f 1 kg1. By the definition of the derivative, f 1 h f 1 f 1 lim kg1 h kg1 lim g1 h g1 lim k c d g1 h g1 k lim c d kg 1 (Factor) (Property of limits) EXAMPLE 3 Applying the constant multiple rule Differentiate the following functions: 4 a. f 1 7 3 b. y 1 3 Solution a. f 1 7 3 b. 4 y 1 3 f 1 7 d d 13 713 1 We conclude this section with the sum and difference rules. dy d 1 d 4 d 14 3 1 a 3 1 1 b 16 3 14 3 18. THE DERIVATIVES OF POLYNOMIAL FUNCTIONS
The Sum Rule If functions p1 and q1 are differentiable, and f 1 p1 q1, then f 1 p 1 q 1. d In Leibniz notation, 1 f 1 d 1p1 d 1q1. d d d Proof: Let f 1 p1 q1. By the definition of the derivative, f 1 h f 1 f 1 lim 3p1 h q1 h4 3p1 q14 lim hs0 h 3p1 h p14 lim e 3p1 h p14 lim e p 1 q 1 3q1 h q14 f h 3q1 h q14 f lim e f The Difference Rule If functions p1 and q1 are differentiable, and f 1 p1 q1, then f 1 p 1 q 1. d In Leibniz notation, 1 f 1 d 1p1 d 1q1. d d d The proof for the difference rule is similar to the proof for the sum rule. EXAMPLE 4 Selecting appropriate rules to determine the derivative Differentiate the following functions: a. f 1 3 5V b. y 13 CHAPTER 19
Solution We apply the constant multiple, power, sum, and difference rules. a. f 1 3 5V b. We first epand y 13. f 1 d d 13 d d 15 1 y 9 1 4 3 d d 1 5 d d 1 1 31 5 a 1 1 b dy 91 111 d 18 1 6 5 or 1 6 5 V EXAMPLE 5 Selecting a strategy to determine the equation of a tangent Determine the equation of the tangent to the graph of f 1 3 3 at 1. Solution A Using the Derivative The slope of the tangent to the graph of f at any point is given by the derivative f 1. For f 1 3 3 f 1 3 6 Now, f 11 311 611 3 6 3 The slope of the tangent at 1 is 3 and the point of tangency is 11, f 11 11, 0. The equation of the tangent is or y 3 3 Tech Support For help using the graphing calculator to graph functions and draw tangent lines See Technical Appendices X XX and X XX. Solution B Using the Graphing Calculator Draw the graph of the function using the graphing calculator. Draw the tangent at the point on the function where 1. The calculator displays the equation of the tangent line. The equation of the tangent line in this case is y 3 3. 0. THE DERIVATIVES OF POLYNOMIAL FUNCTIONS
EXAMPLE 6 Connecting the derivative to horizontal tangents Determine points on the graph in Eample 5 where the tangents are horizontal. Solution Horizontal lines have slope zero. We need to find the values of that satisfy f 1 0. 3 6 0 31 0 0 or The graph of f 1 3 3 has horizontal tangents at 10, and 1,. IN SUMMARY Key Ideas The following table summarizes the derivative rules in this section. Rule Function Notation Leibniz Notation Constant Function Rule If f1 k, where k is a constant, then f 1 0. d 1k 0 d Power Rule If f1 n, where n is a real number, then f 1 n n 1. Constant Multiple Rule If f1 kg1, then f 1 kg 1. Sum Rule If f1 p1 q1, then f 1 p 1 q 1. Difference Rule If f1 p1 q1, then f 1 p 1 q 1. d d 1n n n 1 d 1ky kdy d d d d 1f1 d d 1p1 d d 1q1 d d 1f1 d d 1p1 d d 1q1 Need to Know To determine the derivative of a simple rational function, such as f1 4, 6 epress the function as a power then use the power rule. If f1 4 6, then f 1 41 6 1 6 1 or 4 7 If you have a radical function such as g1 V 3 5, rewrite the function as 5 g1 3 then use the power rule. 5 If, then or 5 3 g 1 5 V 3 3 g1 3 3 CHAPTER 1
Eercise. PART A 1. What rules do you know for calculating derivatives? Give eamples of each rule.. Determine f 1 for each of the following functions: a. f 1 4 7 c. f 1 5 8 e. b. f 1 3 d. f 1 V 3 f. f 1 a b 4 f 1 3 K 3. Differentiate each function. Use either Leibniz notation or prime notation, depending on which is appropriate. a. h1 1 31 4 d. y 1 5 5 1 3 3 1 1 b. f 1 3 5 4 3.75 e. c. s t 1t t f. g1 51 4 4. Apply the differentiation rules you learned in this section to find the derivatives of the following functions: a. y 3 5 c. y 6 e. 3 3 3 s1t t5 3t, t 7 0 t y V 6V 3 V 6 b. y 4 1 d. y 9 3V f. y 1 V PART B 5. Let s represent the position of a moving object at time t. Find the velocity v ds at time t. dt a. s t 7t b. s 18 5t 1 c. s 1t 3 3 t3 6. Determine f 1a for the given function f 1 at the given value of a. a. f 1 3 V, a 4 b. f 1 7 6V 5 3, a 64 7. Determine the slope of the tangent to each of the curves at the given point. a. y 3 4, 11, 3 c. y, 1, 1 b. y 1, 1 1, 1 d. y V16 3, 14, 3 5. THE DERIVATIVES OF POLYNOMIAL FUNCTIONS
8. Determine the slope of the tangent to the graph of each function at the point with the given -coordinate. a. y 3 3, 1 c. y 16, b. y V 5, 4 d. y 3 1 1 1, 1 9. Write an equation of the tangent to each curve at the given point. a. y 1, P10.5, 1 d. y 1 a 1 b, P11, C T A b. y 3 e. 4 3, P1 1, 7 c. y V3 3, P13, 9 f. 10. What is a normal to the graph of a function? Determine the equation of the normal to the graph of the function in question 9, part b, at the given point. 11. Determine the values of so that the tangent to the function y 3 3 is parallel V to the line 16y 3 0. 1. Do the functions y 1 and y ever have the same slope? If so, where? 3 13. Tangents are drawn to the parabola y at 1, 4 and Q 1. Prove that 8, 1 64 R these lines are perpendicular. Illustrate with a sketch. 14. Determine the point on the parabola y 3 4 where the slope of the tangent is 5. Illustrate your answer with a sketch. 15. Determine the coordinates of the points on the graph of y 3 at which the slope of the tangent is 1. 16. Show that there are two tangents to the curve y 1 that have a 5 5 10 slope of 6. 17. Determine the equations of the tangents to the curve y 3 that pass through the following points: a. point 1, 3 b. point 1, 7 18. Determine the value of a, given that the line a 4y 1 0 is tangent to the graph of y a at. 19. It can be shown that, from a height of h metres, a person can see a distance of d kilometres to the horizon, where d 3.53Vh. a. When the elevator of the CN Tower passes the 00 m height, how far can the passengers in the elevator see across Lake Ontario? b. Find the rate of change of this distance with respect to height when the height of the elevator is 00 m. y y 1 V 13V 8, P14, 0 V, P11, 1 3 V CHAPTER 3
0. An object drops from a cliff that is 150 m high. The distance, d, in metres that the object has dropped at t seconds in modelled by d1t 4.9t. a. Find the average rate of change of distance with respect to time from s to 5 s. b. Find the instantaneous rate of change of distance with respect to time at 4 s. c. Find the rate at which the object hits the ground to the nearest tenth. 1. A subway train travels from one station to the net in min. Its distance, in kilometres, from the first station after t minutes is s1t t 1 At what 3 t3. times will the train have a velocity of 0.5 km>min?. While working on a high-rise building, a construction worker drops a bolt from 30 m above the ground. After t seconds, the bolt has fallen a distance of s metres, where s1t 5t, 0 t 8. The function that gives the height of the bolt above ground at time t is R1t 30 5t. Use this function to determine the velocity of the bolt at t. 3. Tangents are drawn from the point 10, 3 to the parabola y 3. Find the coordinates of the points at which these tangents touch the curve. Illustrate your answer with a sketch. 4. The tangent to the cubic function that is defined by y 3 6 8 at point A13, 3 intersects the curve at another point, B. Find the coordinates of B. Illustrate with a sketch. 5. a. Find coordinates of the points, if any, where each function has a horizontal tangent line. i. f 1 5 ii. f 1 4 3 iii. f 1 3 8 5 3 b. Suggest a graphical interpretation for each of these points. PART C 6. Let P1a, b be a point on the curve V Vy 1. Show that the slope of b the tangent at P is Å. a 7. For the power function f 1 n, find the -intercept of the tangent to its graph at point 11, 1. What happens to the -intercept as n increases without bound 1n S q? Eplain the result geometrically. 8. For each function, sketch the graph of y f 1 and find an epression for f 1. Indicate any points at which f 1 does not eist. a. f 1 e, 6 3 b. f 1 0 3 6 0 c. f 1 000 1 0 6, 3 4. THE DERIVATIVES OF POLYNOMIAL FUNCTIONS
Section.3 The Product Rule In this section, we will develop a rule for differentiating the product of two functions, such as f 1 13 11 3 8 and g1 1 3 3 1, without first epanding the epressions. You might suspect that the derivative of a product of two functions is simply the product of the separate derivatives. An eample shows that this is not so. EXAMPLE 1 Reasoning about the derivative of a product of two functions Let k1 f 1g1, where f 1 1 and g1 1 5. Show that k 1 f 1g 1. Solution The epression k1 can be simplified. k1 1 1 5 3 5 10 k 1 3 10 f 1 and g 1 1, so f 1g 1 111. Since is not the derivative of k1, we have shown that k 1 f 1g 1. The correct method for differentiating a product of two functions uses the following rule. The Product Rule If k1 f 1g1, then k 1 f 1g1 f 1g 1. If u and v are functions of, d du dv 1uv v u d d d. In words, the product rule says, the derivative of the product of two functions is equal to the derivative of the first function times the second function plus the first function times the derivative of the second function. Proof: k1 f 1g1, so using the definition of the derivative, k 1 lim hs0 f 1 hg1 h f 1g1. h CHAPTER 5
To evaluate k 1, we subtract and add the same term in the numerator. Now, f 1 hg1 h f 1g1 h f 1g1 h f 1g1 k 1 lim hs0 h f 1 h f 1 g1 h g1 lim ec d g1 h f 1c df h f 1 h f 1 lim c f 1g1 f 1g 1 d limg 1 h hs0 lim hs0 g1 h g1 f 1 lim c d EXAMPLE Applying the product rule Differentiate h1 1 31 5 Solution h1 1 31 5 Using the product rule, we get using the product rule. h 1 d d 3 34 1 5 1 3 1 31 5 1 315 4 6 3 5 4 6 5 6 15 5 7 6 18 5 4 6 d d 35 4 We can, of course, differentiate the function after we first epand. The product rule will be essential, however, when we work with products of polynomials such as f 1 1 91 3 5 4 or non-polynomial functions such as f 1 1 9 V 3 5. It is not necessary to simplify an epression when you are asked to calculate the derivative at a particular value of. Because many epressions derived using differentiation rules are cumbersome, it is easier to substitute, then evaluate the derivative epression. The net eample could be solved by finding the product of the two polynomials and then calculating the derivative of the resulting polynomial at 1. Instead, we will apply the product rule. 6.3 THE PRODUCT RULE
EXAMPLE 3 Selecting an efficient strategy to determine the value of the derivative Find the value h 1 1 for the function h1 15 3 7 31 6. Solution h1 15 3 7 31 6 Using the product rule, we get h 1 115 141 6 15 3 7 314 1 h'1 1 3151 1 141 1431 1 1 1 64 8 351 1 3 71 1 34341 1 14 1117 151 3 The following eample illustrates the etension of the product rule to more than two functions. EXAMPLE 4 Connecting the product rule to a more comple function Find an epression for p 1 if p1 f 1g1h1. Solution We temporarily regard f 1g1 as a single function. p1 3 f 1g14h1 By the product rule, p 1 3 f 1g14 h1 3 f 1g14h 1 A second application of the product rule yields p 1 3 f 1g1 f 1g 14h1 f 1g1h 1 f 1g1h1 f 1g 1h1 f 1g1h 1 This epression gives us the etended product rule for the derivative of a product of three functions. Its symmetrical form makes it easy to etend to a product of four or more functions. The Power of a Function Rule for Positive Integers Suppose that we now wish to differentiate functions such as y 1 3 4 or y 1 3 5 6. CHAPTER 7
These functions are of the form y u n, where n is a positive integer and u g1 is a function whose derivative we can find. Using the product rule, we can develop an efficient method for differentiating such functions. For n, h1 3g14 h1 g1g1 Using the product rule, h 1 g 1g1 g1g 1 g 1g1 Similarly, for n 3, we can use the etended product rule. Thus, h1 3g14 3 g1g1g1 h 1 g 1g1g1 g1g 1g1 g1g1g 1 33g14 g 1 These results suggest a generalization of the power rule. The Power of a Function Rule for Positive Integers If u is a function of, and n d is a positive integer, then d 1un nu n 1 du d. In function notation, if f 1 3g14 n, then f 1 n3g14 n 1 g 1. The power of a function rule is a special case of the chain rule, which we will discuss later in this chapter. We are now able to differentiate any polynomial, such as h1 1 3 5 6 or h1 11 4 1 6 3, without multiplying out 5 the brackets. We can also differentiate rational functions, such as f 1 3 1. EXAMPLE 5 Applying the power of a function rule For h1 1 3 5 6, find h 1. Solution Here h1 has the form h1 3g14 6, where the inner function is g1 3 5. By the power of a function rule, we get h 1 61 3 5 5 1 3. 8.3 THE PRODUCT RULE
EXAMPLE 6 Selecting a strategy to determine the derivative of a rational function 5 Differentiate the rational function f 1 by first epressing it as a 3 1 product and then using the product rule. Solution 5 f 1 3 1 1 513 1 1 (Epress f as a product) f 1 d d 1 513 1 1 1 5 d 13 1 1 d (Product rule) 13 1 1 1 51 113 1 d 13 1 d 13 1 1 31 513 1 31 5 13 1 13 1 13 1 6 15 13 1 13 1 6 6 15 13 1 17 13 1 (Power of a function rule) (Simplify) EXAMPLE 7 Using the derivative to solve a problem The position s, in centimetres, of an object moving in a straight line is given by s t16 3t 4, t 0, where t is the time in seconds. Determine the object s velocity at time t. Solution The velocity of the object at any time v d dt 3t16 3t4 4 t 0 is v ds dt. 1116 3t 4 1t d dt 316 3t4 4 16 3t 4 1t3416 3t 3 1 34 At t, v 0 134101 34 0 We conclude that the object is at rest at time t s. (Product rule) (Power of a function rule) CHAPTER 9
IN SUMMARY Key Ideas The derivative of a product of differentiable functions is not the product of their derivatives. The product rule for differentiation: If h1 f1g1, then h 1 f 1g1 f1g 1. The power of a function rule for positive integers: If f1 3g14 n, then f 1 n3g14 n 1 g 1. Need to Know In some cases, it is easier to epand and simplify the product before differentiating, rather than using the product rule. If f1 3 4 15 3 7 15 7 1 4 f 1 105 6 84 3 If the derivative is needed at a particular value of the independent variable, it is not necessary to simplify before substituting. Eercise.3 K PART A 1. Use the product rule to differentiate each function. Simplify your answers. a. h1 1 4 d. h1 15 7 11 b. h1 1 1 e. s1t 1t 113 t c. h1 13 1 7 f. f 1 3 3. Use the product rule and the power of a function rule to differentiate the following functions. Do not simplify. a. y 15 1 3 1 4 c. y 11 4 1 6 3 b. y 13 413 3 5 d. y 1 9 4 1 1 3 3. When is it not appropriate to use the product rule? Give eamples. 4. Let F1 3b143c14. Epress F 1 in terms of b1 and c1. 30.3 THE PRODUCT RULE
PART B dy 5. Determine the value of for the given value of. d a. y 1 71 3, b. c. d. e. f. y 11 11, 1 y 13 1, y 3 13 7, y 1 1 5 13 4, 1 y 15 15, 3 6. Determine the equation of the tangent to the curve y 1 3 5 13 at the point 11,. 7. Determine the point(s) where the tangent to the curve is horizontal. a. y 1 91 1 b. y 1 11 1 8. Use the etended product rule to differentiate the following functions. Do not simplify. a. y 1 1 3 1 41 3 b. y 13 4 13 3 4 A C T 9. A 75-L gas tank has a leak. After t hours, the remaining volume, V, in litres is V1t 75 a 1 t Use the product rule to determine how 4 b, 0 t 4. quickly the gas is leaking from the tank when the tank is 60% full of gas. 10. Determine the slope of the tangent to h1 1 1 3 1 1 at. Eplain how to find the equation of the normal at. PART C 11. a. Determine an epression for f 1 if f 1 g 1 1g 1g 3 1 p g n 1 1g n 1. b. If f 1 11 11 11 3 p 11 n, find f 10. 1. Determine a quadratic function f 1 a b c if its graph passes through the point 1, 19 and it has a horizontal tangent at 1 1, 8. 13. Sketch the graph of f 1 0 1 0. a. For what values of is f not differentiable? b. Find a formula for f, and sketch the graph of f. c. Find f () at, 0, and 3. 14. Show that the line 4 y 11 0 is tangent to the curve y 16 1. CHAPTER 31
Mid-Chapter Review 1. a. Sketch the graph of f 1 5. b. Calculate the slopes of the tangents to f 1 5 at points with -coordinates 0, 1,,..., 5. c. Sketch the graph of the derivative function f 1. d. Compare the graphs of f 1 and f 1.. Use the definition of the derivative to find f 1 for each function. a. f 1 6 15 c. f 1 5 5 b. f 1 4 d. f 1 V 3. a. Determine the equation of the tangent to the curve y 4 3 at 1. b. Sketch the graph of the function and the tangent. 4. Differentiate each of the following functions: a. y 6 4 c. g1 e. y 111t 1 3 b. y 10 1 d. y 5 3 f. y 1 5. Determine the equation of the tangent to the graph of f 1 4 that has slope 1. 6. Determine f 1 for each of the following functions: a. f 1 4 7 8 d. f 1 V V 3 b. f 1 3 4 5 6 e. f 1 7 3V c. f 1 5 3 3 f. f 1 4 1 5 1 7. Determine the equation of the tangent to the graph of each function. a. y 3 6 4 when 1 b. y 3 V when 9 c. f 1 4 4 3 8 9 when 3 8. Determine the derivative using the product rule. a. f 1 14 913 5 c. y 13 4 61 9 b. f 1t 1 3t 7t 814t 1 d. y 13 3 3 3 MID-CHAPTER REVIEW
9. Determine the equation of the tangent to y 15 9 1 3 at 11, 48. 10. Determine the point(s) where the tangent to the curve y 1 115 is horizontal. dy 11. If y 5 8 4, determine from first principles. d 1. A tank holds 500 L of liquid, which takes 90 min to drain from a hole in the bottom of the tank. The volume, V, remaining in the tank after t minutes is V1t 500 a 1 t 90 b, where 0 t 90 a. How much liquid remains in the tank at 1 h? b. What is the average rate of change of volume with respect to time from 0 min to 60 min? c. How fast is the liquid draining at 30 min? 13. The volume of a sphere is given by V1r 4. 3 pr3 a. Determine the average rate of change of volume with respect to radius as the radius changes from 10 cm to 15 cm. b. Determine the rate of change of volume when the radius is 8 cm. 14. A classmate says, The derivative of a cubic polynomial function is a quadratic polynomial function. Is the statement always true, sometimes true, or never true? Defend your choice in words, and provide two eamples to support your argument. dy 15. Show that and a and b are integers. d 1a 4ba 4b 1 if y a 3b a b 16. a. Determine f 13, where f 1 6 3 4 5 10. b. Give two interpretations of the meaning of f 13. 17. The population, P, of a bacteria colony at t hours can be modelled by P1t 100 10t 10t t 3 a. What is the initial population of the bacteria colony? b. What is the population of the colony at 5 h? c. What is the growth rate of the colony at 5 h? 18. The relative percentage of carbon dioide, C, in a carbonated soft drink at t minutes can be modelled by C1t 100, where t 7. Determine C 1t t and interpret the results at 5 min, 50 min, and 100 min. Eplain what is happening. CHAPTER 33
Section.4 The Quotient Rule In the previous section, we found that the derivative of the product of two functions is not the product of their derivatives. The quotient rule gives the derivative of a function that is the quotient of two functions. It is derived from the product rule. The Quotient Rule If h1 f 1 then h 1 g1, f 1g1 f 1g 1 3g14, du dv In Leibniz notation, d Q u d v R v u d d v. g1 0. Proof: We want to find h 1, given that h1 f 1, g1 0. g1 We rewrite this as a product: h1g1 f 1. Using the product rule, h 1g1 h1g 1 f 1. f 1 h1g 1 Solving for h 1, we get h 1 g1 f 1 f 1 g1 g 1 g1 f 1g1 f 1g 1 3g14 The quotient rule enables us to differentiate rational functions. Memory Aid for the Product and Quotient Rules It is worth noting that the quotient rule is similar to the product rule in that both have f 1g1 and f 1g 1. For the product rule, we put an addition sign between the terms. For the quotient rule, we put a subtraction sign between the terms and then divide the result by the square of the original denominator. 34.4 THE QUOTIENT RULE
EXAMPLE 1 Applying the quotient rule 3 4 Determine the derivative of h1 5. Solution Since h1 f 1 where f 1 3 4 and g1 5, use the quotient g1, rule to find h 1. f 1g1 f 1g 1 Using the quotient rule, we get h 1 3g14 131 5 13 41 1 5 3 15 6 8 1 5 3 8 15 1 5 EXAMPLE Tech Support For help using the graphing calculator to graph functions and draw tangent lines see Technical Appendices X XX and X XX. Selecting a strategy to determine the equation of a line tangent to a rational function Determine the equation of the tangent to y at 0. 1 Solution A Using the Derivative dy The slope of the tangent to the graph of f at any point is given by the derivative d. By the quotient rule, dy d 11 1 11 1 1 At 0, dy 110 1 1010 d 10 1. The slope of the tangent at 0 is and the point of tangency is 10, 0. The equation of the tangent is y. Solution B Using the Graphing Calculator Draw the graph of the function using the graphing calculator. Draw the tangent at the point on the function where 0. The calculator displays the equation of the tangent line. The equation of the tangent line in this case is y. CHAPTER 35
EXAMPLE 3 Using the quotient rule to solve a problem 8 Determine the coordinates of each point on the graph of f1 where the tangent is horizontal. V Solution The slope of the tangent at any point on the graph is given by f 1. Using the quotient rule, 11V 1 8 a 1 1 b f 1 1 V 8 V V 4 V V 4 4 V The tangent will be horizontal when f 1 0; that is, when 4. The point on the graph where the tangent is horizontal is 14, 8. IN SUMMARY Key Ideas The derivative of a quotient of two differentiable functions is not the quotient of their derivatives. The quotient rule for differentiation: If V h1 f1 g1, then h 1 f 1g1 f 1g 1 3g14, g1 0. Need to Know To find the derivative of a rational function, you can use two methods: Leave the function in fraction form, OR Epress the function as a product, and use the quotient rule. and use the product rule. f1 1 f1 1 11 1 36.4 THE QUOTIENT RULE
Eercise.4 PART A 1. What are the eponent rules? Give eamples of each rule.. Copy the table, and complete it without using the quotient rule. Function Rewrite Differentiate and Simplify, If Necessary f1 3, 0 g1 3 5 3, 0 h1 1 105, 0 y 83 6, 0 s t 9 t 3, t 3 C K T 3. What are the different ways to find the derivative of a rational function? Give eamples. PART B 4. Use the quotient rule to differentiate each function. Simplify your answers. a. 13 5 h1 c. h1 3 e. y 1 1 1 b. t 3 d. f. y 1 h1t h1 1 t 5 3 3 dy 5. Determine at the given value of. d a. 3 y 5, 3 c. y 5 5, b. y 3 9, 1 d. 1 11 y 1 11, 4 6. Determine the slope of the tangent to the curve y 3 at point 13, 9. 6 3 7. Determine the points on the graph of y where the slope of the tangent is 1 4 5. 5 8. Show that there are no tangents to the graph of f 1 that have a negative slope. CHAPTER 37
A 9. Find the point(s) at which the tangent to the curve is horizontal. a. b. y 1 y 4 10. An initial population, p, of 1000 bacteria grows in number according to the equation p1t 1000Q1 4t where t is in hours. Find the rate at t 50 R, which the population is growing after 1 h and after h. 11. Determine the equation of the tangent to the curve y 1 at. 3 1. A motorboat coasts toward a dock with its engine off. Its distance s, in metres, from the dock t seconds after the engine is turned off is s1t 10 16 t for 0 t 6. t 3 a. How far is the boat from the dock initially? b. Find the velocity of the boat when it bumps into the dock. 13. The radius of a circular juice blot on a piece of paper towel t seconds after it 1 t was first seen is modelled by r1t where r is measured in 1 t, centimeters. Calculate a. the radius of the blot when it was first observed b. the time at which the radius of the blot was 1.5 cm c. the rate of increase of the area of the blot when the radius was 1.5 cm d. According to this model, will the radius of the blot ever reach cm? Eplain your answer. a b 14. The graph of f 1 has a horizontal tangent line at 1, 1. 1 11 4 Find a and b. Check using a graphing calculator. 15. The concentration, c, of a drug in the blood t hours after the drug is taken 5t orally is given by c1t. When does the concentration reach its t 7 maimum value? 16. The position, s, of an object that moves in a straight line at time t is given by s1t t Determine when the object changes direction. t 8. PART C a b 17. Consider the function f 1 where a, b, c, and d are c d, d c, nonzero constants. What condition on a, b, c, and d ensures that each tangent to the graph of f has a positive slope? 38.4 THE QUOTIENT RULE
Section.5 The Derivatives of Composite Functions Recall that one way of combining functions is through a process called composition. We start with a number in the domain of g, find its image g1, and then take the value of f at g1, provided that g1 is in the domain of f. The result is the new function h1 f 1g1, which is called the composite function of f and g, and is denoted 1 f g. Given two functions f and g, the composite function 1 f g 1 f g1 f 1g1. is defined by EXAMPLE 1 Reflecting of the process of composition If f 1 V and g1 5, find each of the following values: a. f 1g14 b. g1 f 14 c. f 1g1 d. g1 f 1 Solution a. Since g14 9, we have f 1g14 f 19 3. b. Since f 14, we have g1 f 14 g1 7. Note: f 1g14 g 1 f 14 c. f 1g1 f 1 5 V 5 d. g1 f 1 g1 V V 5 Note: f 1g1 g1 f 1 The chain rule states how to compute the derivative of the composite function h1 f 1g1 in terms of the derivatives of f and g. The Chain Rule If f and g are functions that have derivatives, then the composite function h1 f 1g1 has a derivative given by h 1 f 1g1g 1. In words, the chain rule says, the derivative of a composite function is the product of the derivative of the outer function evaluated at the inner function and the derivative of the inner function. CHAPTER 39
Proof: f 1g1 h f 1g1 By the definition of the derivative, 3 f 1g1 4 lim. hs0 h Assuming that g1 h g1 0, we can write f 1g1 h f 1g1 g1 h g1 3 f 1g1 4 lim ca ba bd hs0 g1 h g1 h f 1g1 h f 1g1 g1 h g1 lim c dclim d hs0 g1 h g1 Since lim 3g1 h g14 0, let g1 h g1 k and k S 0 hs0 as h S 0. We obtain f 1g1 k f 1g1 3 f 1g1 4 lim c ks0 k Therefore, 3 f 1g1 4 f 1g1g 1. g1 h g1 dclim (Property of limits) This proof is not valid for all circumstances. When dividing by g1 h g1, we assume that g1 h g1 0. A more advanced proof can be found in advanced calculus tets. d EXAMPLE Applying the chain rule Differentiate h1 1 3. Solution The inner function is g1, and the outer function is f 1 3. The derivative of the inner function is g 1 1. The derivative of the outer function is f 1 3 1. The derivative of the outer function evaluated at the inner function g1 f 1 3 1 1. is By the chain rule, h 1 3 1 1 1 1. The Chain Rule in Leibniz Notation dy dy du dy du If y is a composite function, then provided that and eist. d du d, du d If we interpret derivatives as rates of change, the chain rule states that if y is a function of through the intermediate variable u, then the rate of change of y 40.5 THE DERIVATIVES OF COMPOSITE FUNCTIONS
with respect to is equal to the product of the rate of change of y with respect to u and the rate of change of u with respect to. EXAMPLE 3 Applying the chain rule using Leibniz notation If y u 3 u 1, where u V, find dy at 4. d Solution Using the chain rule, dy dy du d du d 13u ca 1 1 bd 13u a 1 V b It is not necessary to write the derivative entirely in terms of. dy When 4, u V4 4 and d 3314 4Q 1 V4 R 146Q1 R 3. EXAMPLE 4 Selecting a strategy involving the chain rule to solve a problem An environmental study of a certain suburban community suggests that the average daily level of carbon monoide in the air can be modelled by the function C1 p V0.5p 17, where C1 p is in parts per million and population p is epressed in thousands. It is estimated that t years from now, the population of the community will be p1t 3.1 0.1t thousand. At what rate will the carbon monoide level be changing with respect to time three years from now? Solution dc We are asked to find the value of when t 3. dt, We can find the rate of change by using the chain rule. Therefore, dc dt dc dp dp dt d10.5p 17 1 dp d13.1 0.1t c 1 10.5p 17 1 10.51pd10.t When t 3, p13 3.1 0.113 4. dc So, dt c 1 10.514 17 1 10.5114 d 10.13 dt 0.4 dc Since the sign of is positive, the carbon monoide level will be increasing at dt the rate of 0.4 parts per million per year. CHAPTER 41
EXAMPLE 5 Using the chain rule to differentiate a power of a function If y 1 5 7, find dy d. Solution The inner function is g1 5, and the outer function is f 1 7. By the chain rule, dy d 71 5 6 1 141 5 6 Eample 5 is a special case of the chain rule in which the outer function is a power function of the form y 3g14 n. This leads to a generalization of the power rule seen earlier. Power of a Function Rule d If n is a real number and then d 1un nu n 1 du u g1, d or d d 3g14n n 3g14 n 1 g 1. EXAMPLE 6 Connecting the derivative to the slope of a tangent Using graphing technology, sketch the graph of the function f 1 8 4. 4 Tech Support For help using the graphing calculator to graph functions and draw tangent lines see Technical Appendices X XX and X XX..5 THE DERIVATIVES OF COMPOSITE FUNCTIONS Find the equation of the tangent at the point 1, 1 on the graph. Solution Using graphing technology, the graph is The slope of the tangent at point 1, 1 is given by f 1. We first write the function as f 1 81 4 1. By the power of a function rule, f 1 81 4 1. The slope at 1, 1 is f 1 814 4 14 3 18 0.5 The equation of the tangent is y 1 1 1 or y 4 0.
EXAMPLE 7 Combining derivative rules to differentiate a comple product Differentiate h1 1 3 4 14 5 3. Epress your answer in a simplified factored form. Solution Here we use the product rule and the chain rule. h 1 d d 31 3 4 4 14 5 3 d d 314 53 4 1 3 4 341 3 3 14 14 5 3 3314 5 144 1 3 4 81 3 3 14 5 3 114 5 1 3 4 41 3 3 14 5 314 5 31 34 41 3 3 14 5 111 10 9 (Product rule) (Chain rule) (Simplify) (Factor) EXAMPLE 8 Combining derivative rules to differentiate a comple quotient 1 Determine the derivative of g1 Q 1 R 10. Solution A Using the product and chain rule There are several approaches to this problem. You could keep the function as it is and use the chain rule and the quotient rule. You could also decompose the function and epress it as g1 11 10 and then apply the quotient rule 11 10, and the chain rule. Here we will epress the function as the product g1 11 10 11 10 and apply the product rule and the chain rule. g 1 d d c11 10 d11 10 11 10 d d c11 10 d (Simplify) (Factor) (Rewrite using positive eponents) Solution B Using the chain and quotient rule In this solution, we will use the chain rule and the quotient rule, where 1 u is the inner function and u 10 is the outer function. 1 g 1 1011 9 111 10 11 10 1 1011 11 1 011 9 11 10 1011 10 11 11 011 9 11 11 311 11 4 011 9 11 11 1 4011 9 11 11 d311 1 10 4 1 d11 d 1 a d 1 b (Chain rule and quotient rule) CHAPTER 43
1 9 10 a 1 b d 1 a d 1 b 9 1 10 a 1 b c 11 1 11 11 d 9 1 10 a 1 b c 3 3 11 d 9 1 10 a 1 b 4 c 11 d 1011 9 11 9 4 11 4011 9 11 11 (Epand) (Simplify) IN SUMMARY Key Idea The chain rule: dy dy du If y is a composite function, then provided d du d, dy du that and eist. du d Therefore, if h1 1f g1, then h 1 f 1g1 g 1 (Function notation) or d3h14 d d3f 1g1 4 d3g14 d3g14 d (Leibniz notation) Need to Know When the outer function is a power function of the form y 3g14 n, we have a special case of the chain rule, called the power of a function rule: d d 3g14n d3g14n d3g14 d3g14 d n3g14 n 1 g 1 44.5 THE DERIVATIVES OF COMPOSITE FUNCTIONS