16.2 The Derivative of x n where n is any Real Number 521 22. f(x)=x(x + 3) 2 23. f(x)=(x − 2)(x + 1) 2 24. f(x)=(3−x) 2 (x − 1) 25. f(x)=e x (x − 3) 3 26. Prove that if f(x)=(x − a) 3 (x − b) where a, b>0and a = b, then f has a point of inflection at x = a. 27. (a) Which of the following are equal to e −x 2 ? Identify all correct answers. i. e −(x)(x) ii. (e −x ) x iii.  1 e x  x iv.  1 e x  2 v. (e −x ) 2 vi. (e −2 ) x vii. e −2x viii. e (−x) 2 (b) Differentiate e −x 2 . 28. Graph f(x)=e −x 2 and answer the following questions. (a) What is the domain of f ? The range? (b) Is f an even function, an odd function, or neither? (c) For what values of x is f increasing? Decreasing? (d) Find all relative maximum and minimum points. (e) Does f have an absolute maximum value? An absolute minimum value? A greatest lower bound? If any of these exist, identify them. (f) Find the x-coordinates of the points of inflection. (Note: This function is a good one to keep in mind. It is extremely useful in both probability and statistics because, with some minor adjustments, it gives us a normal distribution curve.) 16.2 THE DERIVATIVE OF x n WHERE n IS ANY REAL NUMBER At this point we have proven that d dx x n = nx n−1 for any integer n. In fact, this formula holds if n is any real number. We can use the Chain Rule to prove this fact. Suppose that r is any real number and we want to find d dx x r . The key is to rewrite x r so we can use derivative formulas that we already know. x r = e ln x r e x and ln x are inverses, so e ln z = z. = e r ln x Use log rule (iii). Now we take the derivative. 522 CHAPTER 16 Taking the Derivative of Composite Functions d dx x r = d dx e r ln x = e r ln x · d dx (r ln x) Use the Chain Rule: d dx e mess = e mess · (mess)  . = e r ln x · r x Keep in mind that r is a constant. = x r · r x Rewrite e r ln x as x r . = rx r−1 We now can take the derivative of x n where n is any real number. d dx x n =nx n−1 EXERCISE 16.1 Find y  if y = x π π x +(2x) 2.1 π 2 . Answer y  = π −2 [πx π−1 π x + x π (ln π) π x + (2.1)2 2.1 x 1.1 ] PROBLEMS FOR SECTION 16.2 For Problems 1 through 3, find y  . 1. y = x 2π + 2π x 2. y = (2x 2 + 1) √ 3 3. y = (3x) √ 2+1 + 1 √ πx 4. Differentiate the following, simplifying the expression first if useful. (a) y = πe 3t 2 +π (b) y = ln(e t + 1) (c) y = π 2 √ x 2 +4 (d) y = 1 (ln x) 2.6 (e) y = 1 (ln x 2 ) 1.5 (f) y = 3 √ ln(e t + 1) 5. Find y  , simplifying the expression first where useful. (a) y = e x x e (b) y = e 1/x (c) y = √ e −x x (d) y =  ln √ 1 − x  −3.5 (e) y = ln  x+1 x−1  (f) y = (1 − ln x) 5/4 (g) y = ln √ x(x +1) (h) y = 5  1 e 6x +x 6. Identify and classify all critical points of the function f(x)=(x 2 − 4)x π+1 for x>0. 16.3 Using the Chain Rule 523 16.3 USING THE CHAIN RULE In this section, we take a second look at applying the Chain Rule. ◆ EXAMPLE 16.6 Once we know the Chain Rule, there is no problem if we forget the Quotient Rule, because f(x) g(x) can be written as f(x)·[g(x)] −1 . Derive the Quotient Rule from the Product Rule. SOLUTION d dx  f(x) g(x)  = d dx  f(x)·[g(x)] −1  Rewrite the quotient as a product. = f(x)· d dx [g(x) −1 ] + [g(x) −1 ] · d dx f(x) Use the Product Rule. = f(x)·(−1)[g(x)] −2 g  (x) + [g(x)] −1 · f  (x) Use the Chain Rule to find d dx [g(x) −1 ] =−1[g(x)] −2 · g  (x). = −f(x)·g  (x) [g(x)] 2 + f  (x) g(x) Rewrite negative exponents as fractions. = −f(x)·g  (x) [g(x)] 2 + g(x)·f  (x) [g(x)] 2 Get a common denominator of [g(x)] 2 . = g(x)·f  (x)−f(x)·g  (x) [g(x)] 2 Combine into one fraction. ◆ We can apply the Chain Rule to relate rates of change; as illustrated in the following example. ◆ EXAMPLE 16.7 In Nepal the daily kerosene consumption in restaurants and guesthouses throughout the country can be modeled by the function K(x), where x is the number of tourists in the country. Tourism is seasonal; let x(t) be the number of tourists at time t. In our model, K(x) and x(t) are continuous and differentiable. At a certain time, there are 5,000 tourists in Nepal and the number is decreasing at a rate of 40 per day. Write an expression for the rate at which kerosene consumption is changing with time at this moment. SOLUTION We’re looking for dK dt . We know that dK dt = dK dx · dx dt , and dx dt =−40. Therefore, dK dt = K  (5000) · (−40) =−40K  (5000), where K  (5000) = dK dx     x=5000 . ◆ Before applying the Chain Rule to more complicated expressions and situations, we’ll summarize our knowledge of differentiation, using the Chain Rule to express familiar differentiation formulas in a more general way. Derivatives: A Summary Graphical Interpretation: If y = f(x),then the derivative of f is the slope function, giving the slope of the tangent line to f at the point (x, f(x)).For instance, f  (2) gives the slope of f at x = 2. Instantaneous Rate of Change: The derivative evaluated at x = 2 gives the instantaneous rate of change of y with respect to x at x = 2. Notation: y  , f  , f  (x), y  (x), dy dx , df dx all mean the same thing. d dx is an operator that means “take the derivative of what follows.” 524 CHAPTER 16 Taking the Derivative of Composite Functions Limit Definition: The following expressions are equivalent and are equal to f  (x). f  (x) = lim h→0 f(x +h) − f(x) h = lim x→0 y x = lim x→0 f(x +x) − f(x) x Differentiation Formulas 1. y = f(x)+g(x) y  = f  (x) + g  (x) Sum Rule 2. y = kf (x ) k constant y  = kf  (x) Constant Multiple Rule 3. y = f(x)g(x) y  = f  (x)g(x) + g  (x)f (x) Product Rule For the product of two functions 4. y = f(x) g(x) y  = gf  − fg  g 2 Quotient Rule 5. y = [g(x)] n n constant y  = n[g(x)] n−1 · g  (x) Generalized Power Rule (Variable in base) 6. y = b g(x) b constant y  = b g(x) (ln b)g  (x) Exponential Function (Variable in exponent) y = e g(x) y  = e g(x) g  (x) 7. y = log b  g(x)  y  = 1 ln b · 1 g(x) g  (x) Logarithmic Function y = ln[g(x)] y  = 1 g(x) g  (x) 8. y = f (g(x)) y  = f  (g(x))g  (x) Chain Rule The world of functions we can differentiate has broadened immensely! If faced with a function we cannot differentiate using these shortcuts, 2 we can return to the limit definition of derivative. If we cannot make headway obtaining an exact value for a derivative and the function we’re working with is reasonably well behaved, we can still use numerical methods to approximate the derivative at a point, and we can obtain qualitative information using a graphical approach. CAUTION Be sure you know the difference between an exponential function like f(x)= (π + 1) 2x and a generalized power function like f(x)= (x 2 + π) π 2 . If the variable is in the base, then we differentiate using the generalized power rule; if the variable is in the exponent, we use the generalized rule for exponential functions. Feel free to think of some of these differentiation formulas more informally. For instance, the derivative of (mess) n is n(mess) n−1 · (mess)  , the derivative of ln(mess)is 1 mess · (mess)  , and the derivative of e mess is e mess · (mess)  . 2 This will be the case when we first run into trigonometric functions. 16.3 Using the Chain Rule 525 These rules are all obtained by applying the Chain Rule to the differentiation rules we have been using. Then when you look at a complicated expression, you can categorize it. For instance, the expression 1  ln(x 2 + 1) is basically (mess) −1/2 , meaning were we to construct the expression assembly-line style, the last worker on the line would take the mess coming along the line and raise it to the −1/2. The derivative of (mess) −1/2 is − 1 2 (mess) −3/2 · (mess)  . What’s the mess? Basically, it is ln(stuff),so (mess)  = 1 stuff · (stuff )  . Putting this together, d dx  1  ln(x 2 + 1)  = d dx  ln(x 2 + 1)  − 1 2 =− 1 2  ln(x 2 + 1)  − 3 2 · 1 (x 2 + 1) · 2x. What we’ve done—albeit informally—is to decompose 1 √ ln(x 2 +1) into f (g(h(x))), where f(x)=x −1/2 , g(x) = ln x, and h(x) = x 2 + 1. x h −→  x 2 + 1  g −→ ln(x 2 + 1) f −→ 1  ln(x 2 + 1) So  f (g(h(x)))   = f  (g(h(x))) · g  (h(x)) · h  (x). A WORD OF ADVICE Many students, having learned shortcuts to differentiation, approach problems by charging them, sleeves rolled up, brutally hacking away. There is no virtue in doing a problem in the most difficult way possible; it does not put you on higher moral ground. Rather, you increase the likelihood of making an error. Instead, when presented with a function to differentiate, take a moment to consider how to prepare the function for differentiation. Sometimes, a bit of thoughtful reformulation will save you time and energy. (Of course, there are times when it is necessary to dig in and get your hands all dirty. If it is necessary, do it with relish!) The meaning of this advice is clarified in the next example. ◆ EXAMPLE 16.8 Differentiate y = ln  1+x (1−x) 3 . SOLUTION If you’re enthusiastically charging the problem, you might informally say that basically this is ln(mess), where mess is (stuff ) 1/2 and to differentiate stuff , the Quotient Rule and generalized power rule come into play. If you’re formally charging headlong into this, you might valiantly decompose the function as follows: 526 CHAPTER 16 Taking the Derivative of Composite Functions h(x) = 1 + x (1 − x) 3 g(x) = √ x f(x)=ln x, where ln  1 + x (1 − x) 3 =f (g(h(x))). Notice that h  alone is a bulky expression. Alternatively, you can save yourself a lot of work by rewriting ln  1+x (1−x) 3 : ln  1 + x (1 − x) 3 =ln  1 + x (1 − x) 3  1 2 = 1 2 ln  1 + x (1 − x) 3  = 1 2 ln(1 + x) − 1 2 ln(1 − x) 3 = 1 2 ln(1 + x) − 3 2 ln(1 − x). Now the expression is ready for differentiation. dy dx =  1 2  1 1 + x −  3 2  1 1 − x ·−1 = 1 2(1 + x) + 3 2(1 − x) = 1 2 + 2x + 3 2 − 2x ◆ ◆ EXAMPLE 16.9 Suppose f and g are differentiable functions. Find h  (x) if h(x) =  [f(x)] 3 ·ln(g(x)). SOLUTION This is basically of the form (mess) 1/2 where mess is the product [f(x)] 3 ·ln(g(x)). h  (x) = 1 2 (mess) −1/2 , (mess)  = 1 2  [f(x)] 3 ·ln(g(x))  −1/2 ·  d dx [f(x)] 3 ·ln(g(x)) + [f(x)] 3 d dx  ln(g(x))   = 1 2  [f(x)] 3 ·ln(g(x))  −1/2 ·  3[f(x)] 2 ·f  (x) · ln(g(x)) + [f(x)] 3 · g  (x) g(x)  Notice that the operator notation, d dx , is useful for record-keeping. It can be read as “I plan to take the derivative of this, but I haven’t done it yet.” ◆ 16.3 Using the Chain Rule 527 ◆ EXAMPLE 16.10 Two ships leave from a port. The first one sails due east, while the second one sails due north. At 10:00 a.m. the first boat is 4 miles from the port and is traveling at 35 miles per hour. At this instant the second boat is 3 miles north of the port and is traveling at 15 miles per hour. At what rate is the distance between the ships increasing at 10:00 a.m.? SOLUTION Let x(t) = the distance between the port and the first boat at time t. y(t) = the distance between the port and the second boat at time t. Then the distance between the two boats at time t, t in hours, is given by D(t) =  [x(t)] 2 +[y(t)] 2 . We want to find dD dt at 10:00 a.m. D(t) is basically of the form (mess) 1/2 . D  (t) = 1 2  [x(t)] 2 +[y(t)] 2  −1/2 ·  2x(t) ·x  (t) + 2y(t) ·y  (t)  At 10:00a.m. x(t) =4, y(t) =3, x  (t) = 35, and y  (t) = 15, so dD dt = 1 2 (4 2 + 3 2 ) −1/2 · [2 · 4 · 35 + 2 · 3 · 15] = 1 2 1 √ 25 ·  2(140 + 45)  = 1 5 · 185 = 37. At 10:00 a.m. the distance between the ships is increasing at 37 miles per hour. ◆ In Section 17.4 we will look at other ways of approaching this problem and problems similar. These other approaches depend, ultimately, on the Chain Rule. 528 CHAPTER 16 Taking the Derivative of Composite Functions Exploratory Problems for Chapter 16 Finding the Best Path 1. A pumping station that will service two towns is to be built on the shore of a straight river. The towns are 10 miles apart; one of them is 3 miles from the river and the other is 9 miles from the river. Pipes will run from the pumping station to the center of each town. Where should the pumping station be located in order to minimize the amount of piping used? town town town town 9 3 10 River L Pumping station Hint: Begin by determining the distance L in the figure above. Then the amount of piping can be expressed in terms of x where x is the distance between the pumping station and the point on the river closest to one of the towns. What is the domain of the function you are minimizing? 2. A family lives on the western shore of a long, straight river in a rural area. The only store in the vicinity is located on the eastern shore of the river, directly across from a point 6 miles down the river. The river is 1 mile wide. They can get to the store by a combination of boat and foot. Where should they build their dock so the commute to the store takes the minimum amount of time? home 6 miles R I V E R 1 mile Store (with store's dock) (a) Answer this question assuming that one can walk at a rate of 4 miles per hour and row at a rate of 3 miles per hour. (b) Now assume that one can walk at a rate of 4 miles per hour, but with a new boat upgrade, the boating rate can be increased to 5 miles per hour. Where should the dock be built if they purchase a new boat? Exploratory Problems for Chapter 16 529 PROBLEMS FOR SECTION 16.3 1. Just outside Newburgh, the New York State Thruway (I-87), running north-south, intersects Interstate 84, which runs east-west. At noon a car is at this intersection and traveling north at a constant speed of 55 miles per hour. At this moment a Greyhound bus is 150 miles west of the intersection and traveling east at a steady pace of 65 miles per hour. (a) When will the bus and the car be closest to one another? (b) What is the minimum distance between the two vehicles? (c) How far away from the intersection is the bus at this time? (Hint: Minimize the square of the distance rather than the distance itself. Explain why this strategy is valid.) 2. Draw a semicircle of radius 2. Inscribe a rectangle as shown. What are the dimensions of the rectangle of the largest area? What is the largest area? (Hint: Strategize. What is your goal? What are you trying to maximize? Write an expression for the thing you’re trying to maximize. Make sure you get it in one variable. Find and classify the critical points.) 3. What is the largest value of ln(x+1) x+1 ? 4. Differentiate. (a) y = 1 x ln 2+1 (b) y = ln(5x 3 + 8x) (c) y = (2 x )(x 2 + x) 7 (d) y =  ln(5x) +e 6x (e) y = 7 √ ln x (f) y =  4 x/3  ln 3x 5. Differentiate. (a) y =  2 2  x (b) y = 2 2 x (c) y = e πx x (d) y = x 3 +1 x 2 +1 (e) y = 5ln  5x+3 √ x  (f) y = 3 2ln(8x 2 +1) 6. What is the global maximum value of the function f(x)= 3 √ x 2 +1 and where is it attained? Instructions: First just look at this function. Without any calculus, try to figure out the answer. (It may be useful to check symmetry considerations.) Now use the first derivative to support your answer. 530 CHAPTER 16 Taking the Derivative of Composite Functions 7. The graph of h(x) is given below. 1 1 –1 –1–2–3 –2 –3 – 4 2 2 3 3 45 (–3, –4) (3, –3) (5, 2) y h(x) x (a) For which values of x is h  (x) negative? (b) This part of the question concerns the function f given by f(x)=[h(x)] 2 , where the graph of h(x) is given above. i. Find all the points (x-values) at which the graph of f(x) has a horizontal tangent line. ii. On a number line, indicate the sign of f  (x). iii. Identify the x-coordinates of all the local maxima and minima of f . iv. On the interval [−3, 5], what is the largest value taken on by f (i.e., what is the global or absolute maximum value of f for x in [−3, 5])? v. On the interval [−3, 5], what is the smallest value taken on by f ? (c) Let g(x) =|h(x)|. i. Sketch the graph of g(x). ii. Is g  (x) defined everywhere? If not, where is g  (x) undefined? 8. Suppose that f and g are differentiable functions. We are given the following informa- tion: x f(x) f  (x) g(x) g  (x) 0 35 2−1 1 2 −102 2 0.3 2.5 4 −2 3 45 1 3 4 11 2 3 5 Evaluate the following. If there is not enough information to do so, indicate this. (a) y = f (g(x)). What is dy dx    x=0 ? (b) y = g(f (x)). What is y(0)? . instantaneous rate of change of y with respect to x at x = 2. Notation: y  , f  , f  (x), y  (x), dy dx , df dx all mean the same thing. d dx is an operator that means “take the derivative of. They can get to the store by a combination of boat and foot. Where should they build their dock so the commute to the store takes the minimum amount of time? home 6 miles R I V E R 1 mile Store. will save you time and energy. (Of course, there are times when it is necessary to dig in and get your hands all dirty. If it is necessary, do it with relish!) The meaning of this advice is clarified