172 We have seen how to interpret derivatives as slopes and rates of change. We have seen how to estimate derivatives of functions given by tables of values. We have learned how to graph derivatives of functions that are defined graphically. We have used the definition of a derivative to calculate the derivatives of functions defined by formulas. But it would be tedious if we always had to use the definition, so in this chapter we develop rules for finding derivatives without having to use the definition directly. These differentiation rules enable us to calculate with relative ease the derivatives of poly- nomials, rational functions, algebraic functions, exponential and logarithmic functions, and trigonometric and inverse trigonometric functions. We then use these rules to solve problems involving rates of change and the approximation of functions. By measuring slopes at points on the sine curve, we get strong visual evidence that the derivative of the sine function is the cosine function. DIFFERENTIATION RULES 3 x ƒ=y= sin x 0 x y y fª(xy= ) 0 pi 2 m=1 m=_1 m=0 pi 2 pi pi DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS In this section we learn how to differentiate constant functions, power functions, polyno- mials, and exponential functions. Let’s start with the simplest of all functions, the constant function . The graph of this function is the horizontal line y H33527 c, which has slope 0, so we must have . (See Figure 1.) A formal proof, from the definition of a derivative, is also easy: In Leibniz notation, we write this rule as follows. DERIVATIVE OF A CONSTANT FUNCTION POWER FUNCTIONS We next look at the functions , where n is a positive integer. If , the graph of is the line y H33527 x, which has slope 1. (See Figure 2.) So (You can also verify Equation 1 from the definition of a derivative.) We have already investigated the cases and . In fact, in Section 2.8 (Exercises 17 and 18) we found that For we find the derivative of as follows: Thus d dx H20849x 4 H20850 H33527 4x 3 3 H33527 lim hl0 H208494x 3 H11001 6x 2 h H11001 4xh 2 H11001 h 3 H20850 H33527 4x 3 H33527 lim hl0 4x 3 h H11001 6x 2 h 2 H11001 4xh 3 H11001 h 4 h H33527 lim hl0 x 4 H11001 4x 3 h H11001 6x 2 h 2 H11001 4xh 3 H11001 h 4 H11002 x 4 h fH11032H20849xH20850 H33527 lim hl0 f H20849x H11001 hH20850 H11002 f H20849xH20850 h H33527 lim hl0 H20849x H11001 hH20850 4 H11002 x 4 h f H20849xH20850 H33527 x 4 n H33527 4 d dx H20849x 3 H20850 H33527 3x 2 d dx H20849x 2 H20850 H33527 2x2 n H33527 3n H33527 2 d dx H20849xH20850 H33527 11 f H20849xH20850 H33527 x n H33527 1f H20849xH20850 H33527 x n d dx H20849cH20850 H33527 0 H33527 lim h l 0 0 H33527 0 fH11032H20849xH20850 H33527 lim hl0 f H20849x H11001 hH20850 H11002 f H20849xH20850 h H33527 lim hl0 c H11002 c h fH11032H20849xH20850 H33527 0 f H20849xH20850 H33527 c 3.1 173 FIGURE 1 The graph of ƒ=c is the line y=c, so fª(x)=0. y c 0 x y=c slope=0 6 030102 y 0 x y=x slope=1 FIGURE 2 The graph of ƒ=x is the line y=x, so fª(x)=1. Comparing the equations in (1), (2), and (3), we see a pattern emerging. It seems to be a reasonable guess that, when n is a positive integer, . This turns out to be true. THE POWER RULE If n is a positive integer, then FIRST PROOF The formula can be verified simply by multiplying out the right-hand side (or by summing the second factor as a geometric series). If , we can use Equation 2.7.5 for and the equation above to write SECOND PROOF In finding the derivative of we had to expand . Here we need to expand and we use the Binomial Theorem to do so: because every term except the first has as a factor and therefore approaches 0. M We illustrate the Power Rule using various notations in Example 1. EXAMPLE 1 (a) If , then . (b) If , then H33527 . (c) If , then . (d) M d dr H20849r 3 H20850 H33527 3r 2 dy dt H33527 4t 3 y H33527 t 4 1000x 999 yH11032y H33527 x 1000 fH11032H20849xH20850 H33527 6x 5 f H20849xH20850 H33527 x 6 h H33527 nx nH110021 H33527 lim hl0 H20875 nx nH110021 H11001 nH20849n H11002 1H20850 2 x nH110022 h H11001 H11080H11080H11080 H11001 nxh nH110022 H11001 h nH110021 H20876 H33527 lim hl0 nx nH110021 h H11001 nH20849n H11002 1H20850 2 x nH110022 h 2 H11001 H11080H11080H11080 H11001 nxh nH110021 H11001 h n h fH11032H20849xH20850 H33527 lim hl0 H20875 x n H11001 nx nH110021 h H11001 nH20849n H11002 1H20850 2 x nH110022 h 2 H11001 H11080H11080H11080 H11001 nxh nH110021 H11001 h n H20876 H11002 x n h H20849x H11001 hH20850 n H20849x H11001 hH20850 4 x 4 fH11032H20849xH20850 H33527 lim hl0 f H20849x H11001 hH20850 H11002 f H20849xH20850 h H33527 lim hl0 H20849x H11001 hH20850 n H11002 x n h H33527 na nH110021 H33527 a nH110021 H11001 a nH110022 a H11001 H11080H11080H11080 H11001 aa nH110022 H11001 a nH110021 H33527 lim xla H20849x nH110021 H11001 x nH110022 a H11001 H11080H11080H11080 H11001 xa nH110022 H11001 a nH110021 H20850 fH11032H20849aH20850 H33527 lim xla f H20849xH20850 H11002 f H20849aH20850 x H11002 a H33527 lim xla x n H11002 a n x H11002 a fH11032H20849aH20850f H20849xH20850 H33527 x n x n H11002 a n H33527 H20849x H11002 aH20850H20849x nH110021 H11001 x nH110022 a H11001 H11080H11080H11080 H11001 xa nH110022 H11001 a nH110021 H20850 d dx H20849x n H20850 H33527 nx nH110021 H20849dH20862dxH20850H20849x n H20850 H33527 nx nH110021 174 |||| CHAPTER 3 DIFFERENTIATION RULES N The Binomial Theorem is given on Reference Page 1. What about power functions with negative integer exponents? In Exercise 61 we ask you to verify from the definition of a derivative that We can rewrite this equation as and so the Power Rule is true when . In fact, we will show in the next section [Exercise 58(c)] that it holds for all negative integers. What if the exponent is a fraction? In Example 3 in Section 2.8 we found that which can be written as This shows that the Power Rule is true even when . In fact, we will show in Sec- tion 3.6 that it is true for all real numbers n. THE POWER RULE (GENERAL VERSION) If n is any real number, then EXAMPLE 2 Differentiate: (a) (b) SOLUTION In each case we rewrite the function as a power of x. (a) Since , we use the Power Rule with : (b) M The Power Rule enables us to find tangent lines without having to resort to the defi- nition of a derivative. It also enables us to find normal lines. The normal line to a curve at a point is the line through that is perpendicular to the tangent line at . (In the study of optics, one needs to consider the angle between a light ray and the normal line to a lens.) PPP C dy dx H33527 d dx (s 3 x 2 ) H33527 d dx H20849x 2H208623 H20850 H33527 2 3 x H208492H208623H20850H110021 H33527 2 3 x H110021H208623 fH11032H20849xH20850 H33527 d dx H20849x H110022 H20850 H33527 H110022x H110022H110021 H33527 H110022x H110023 H33527 H11002 2 x 3 n H33527 H110022f H20849xH20850 H33527 x H110022 y H33527s 3 x 2 f H20849xH20850 H33527 1 x 2 d dx H20849x n H20850 H33527 nx nH110021 n H33527 1 2 d dx H20849x 1H208622 H20850 H33527 1 2 x H110021H208622 d dx sx H33527 1 2sx n H33527 H110021 d dx H20849x H110021 H20850 H33527 H20849H110021H20850x H110022 d dx H20873 1 x H20874 H33527 H11002 1 x 2 SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS |||| 175 2 _2 _3 3 y yª FIGURE 3 y=#œ„≈ N Figure 3 shows the function in Example 2(b) and its derivative . Notice that is not differ- entiable at ( is not defined there). Observe that is positive when increases and is nega- tive when decreases.y yyH11032 yH110320 yyH11032 y EXAMPLE 3 Find equations of the tangent line and normal line to the curve at the point . Illustrate by graphing the curve and these lines. SOLUTION The derivative of is So the slope of the tangent line at (1, 1) is . Therefore an equation of the tan- gent line is The normal line is perpendicular to the tangent line, so its slope is the negative recipro- cal of , that is, . Thus an equation of the normal line is We graph the curve and its tangent line and normal line in Figure 4. M NEW DERIVATIVES FROM OLD When new functions are formed from old functions by addition, subtraction, or multipli- cation by a constant, their derivatives can be calculated in terms of derivatives of the old functions. In particular, the following formula says that the derivative of a constant times a function is the constant times the derivative of the function. THE CONSTANT MULTIPLE RULE If c is a constant and is a differentiable func- tion, then PROOF Let . Then (by Law 3 of limits) M EXAMPLE 4 (a) (b) M The next rule tells us that the derivative of a sum of functions is the sum of the derivatives. d dx H20849H11002xH20850 H33527 d dx H20851H20849H110021H20850xH20852 H33527 H20849H110021H20850 d dx H20849xH20850 H33527 H110021H208491H20850 H33527 H110021 d dx H208493x 4 H20850 H33527 3 d dx H20849x 4 H20850 H33527 3H208494x 3 H20850 H33527 12x 3 H33527 cfH11032H20849xH20850 H33527 c lim hl0 f H20849x H11001 hH20850 H11002 f H20849xH20850 h H33527 lim hl0 c H20875 f H20849x H11001 hH20850 H11002 f H20849xH20850 h H20876 tH11032H20849xH20850 H33527 lim hl0 tH20849x H11001 hH20850 H11002tH20849xH20850 h H33527 lim hl0 cf H20849x H11001 hH20850 H11002 cf H20849xH20850 h tH20849xH20850 H33527 cf H20849xH20850 d dx H20851cf H20849xH20850H20852 H33527 c d dx f H20849xH20850 f y H33527 H11002 2 3 x H11001 5 3 ory H11002 1 H33527 H11002 2 3 H20849x H11002 1H20850 H11002 2 3 3 2 y H33527 3 2 x H11002 1 2 ory H11002 1 H33527 3 2 H20849x H11002 1H20850 fH11032H208491H20850 H33527 3 2 fH11032H20849xH20850 H33527 3 2 x H208493H208622H20850H110021 H33527 3 2 x 1H208622 H33527 3 2 sx f H20849xH20850 H33527 xsx H33527 xx 1H208622 H33527 x 3H208622 H208491, 1H20850 y H33527 xsx V 176 |||| CHAPTER 3 DIFFERENTIATION RULES 3 _1 _1 3 tangent normal FIGURE 4 N GEOMETRIC INTERPRETATION OF THE CONSTANT MULTIPLE RULE x y 0 y=2ƒ y=ƒ Multiplying by stretches the graph verti- cally by a factor of 2. All the rises have been doubled but the runs stay the same. So the slopes are doubled, too. c H33527 2 THE SUM RULE If f and t are both differentiable, then PROOF Let . Then (by Law 1) M The Sum Rule can be extended to the sum of any number of functions. For instance, using this theorem twice, we get By writing as and applying the Sum Rule and the Constant Multiple Rule, we get the following formula. THE DIFFERENCE RULE If f and t are both differentiable, then The Constant Multiple Rule, the Sum Rule, and the Difference Rule can be combined with the Power Rule to differentiate any polynomial, as the following examples demonstrate. EXAMPLE 5 M H33527 8x 7 H11001 60x 4 H11002 16x 3 H11001 30x 2 H11002 6 H33527 8x 7 H11001 12H208495x 4 H20850 H11002 4H208494x 3 H20850 H11001 10H208493x 2 H20850 H11002 6H208491H20850 H11001 0 H33527 d dx H20849x 8 H20850 H11001 12 d dx H20849x 5 H20850 H11002 4 d dx H20849x 4 H20850 H11001 10 d dx H20849x 3 H20850 H11002 6 d dx H20849xH20850 H11001 d dx H208495H20850 d dx H20849x 8 H11001 12x 5 H11002 4x 4 H11001 10x 3 H11002 6x H11001 5H20850 d dx H20851 f H20849xH20850 H11002tH20849xH20850H20852 H33527 d dx f H20849xH20850 H11002 d dx tH20849xH20850 f H11001 H20849H110021H20850tf H11002t H20849 f H11001tH11001 hH20850H11032 H33527 H20851H20849 f H11001tH20850 H11001 hH20852H11032 H33527 H20849 f H11001tH20850H11032 H11001 hH11032 H33527 fH11032 H11001tH11032 H11001 hH11032 H33527 fH11032H20849xH20850 H11001tH11032H20849xH20850 H33527 lim hl0 f H20849x H11001 hH20850 H11002 f H20849xH20850 h H11001 lim hl0 tH20849x H11001 hH20850 H11002tH20849xH20850 h H33527 lim hl0 H20875 f H20849x H11001 hH20850 H11002 f H20849xH20850 h H11001 tH20849x H11001 hH20850 H11002tH20849xH20850 h H20876 H33527 lim hl0 H20851 f H20849x H11001 hH20850 H11001tH20849x H11001 hH20850H20852 H11002 H20851 f H20849xH20850 H11001tH20849xH20850H20852 h FH11032H20849xH20850 H33527 lim hl0 FH20849x H11001 hH20850 H11002 FH20849xH20850 h FH20849xH20850 H33527 f H20849xH20850 H11001tH20849xH20850 d dx H20851 f H20849xH20850 H11001tH20849xH20850H20852 H33527 d dx f H20849xH20850 H11001 d dx tH20849xH20850 SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS |||| 177 N Using prime notation, we can write the Sum Rule as H20849 f H11001tH20850H11032 H33527 fH11032 H11001tH11032 EXAMPLE 6 Find the points on the curve where the tangent line is horizontal. SOLUTION Horizontal tangents occur where the derivative is zero. We have Thus if x H33527 0 or , that is, . So the given curve has horizontal tangents when x H33527 0, , and . The corresponding points are , , and . (See Figure 5.) M EXAMPLE 7 The equation of motion of a particle is , where is measured in centimeters and in seconds. Find the acceleration as a function of time. What is the acceleration after 2 seconds? SOLUTION The velocity and acceleration are The acceleration after 2 s is . M EXPONENTIAL FUNCTIONS Let’s try to compute the derivative of the exponential function using the defini- tion of a derivative: The factor doesn’t depend on h, so we can take it in front of the limit: Notice that the limit is the value of the derivative of at , that is, Therefore we have shown that if the exponential function is differentiable at 0, then it is differentiable everywhere and This equation says that the rate of change of any exponential function is proportional to the function itself. (The slope is proportional to the height.) fH11032H20849xH20850 H33527 fH11032H208490H20850a x 4 f H20849xH20850 H33527 a x lim hl0 a h H11002 1 h H33527 fH11032H208490H20850 0f fH11032H20849xH20850 H33527 a x lim hl0 a h H11002 1 h a x H33527 lim h l 0 a x a h H11002 a x h H33527 lim h l 0 a x H20849a h H11002 1H20850 h fH11032H20849xH20850 H33527 lim h l 0 f H20849x H11001 hH20850 H11002 f H20849xH20850 h H33527 lim h l 0 a xH11001h H11002 a x h f H20849xH20850 H33527 a x aH208492H20850 H33527 14 cmH20862s 2 aH20849tH20850 H33527 dv dt H33527 12t H11002 10 vH20849tH20850 H33527 ds dt H33527 6t 2 H11002 10t H11001 3 t ss H33527 2t 3 H11002 5t 2 H11001 3t H11001 4 (H11002s3, H110025)(s3, H110025) H208490, 4H20850H11002s3s3 x H33527 H11006s3x 2 H11002 3 H33527 0dyH20862dx H33527 0 H33527 4x 3 H11002 12x H11001 0 H33527 4xH20849x 2 H11002 3H20850 dy dx H33527 d dx H20849x 4 H20850 H11002 6 d dx H20849x 2 H20850 H11001 d dx H208494H20850 y H33527 x 4 H11002 6x 2 H11001 4V 178 |||| CHAPTER 3 DIFFERENTIATION RULES FIGURE 5 The curve y=x$-6x@+4 and its horizontal tangents 0 x y (0, 4) {œ„3, _5}{_œ„3, _5} Numerical evidence for the existence of is given in the table at the left for the cases and . (Values are stated correct to four decimal places.) It appears that the limits exist and In fact, it can be proved that these limits exist and, correct to six decimal places, the val- ues are Thus, from Equation 4, we have Of all possible choices for the base in Equation 4, the simplest differentiation formula occurs when . In view of the estimates of for and , it seems rea- sonable that there is a number between 2 and 3 for which . It is traditional to denote this value by the letter . (In fact, that is how we introduced e in Section 1.5.) Thus we have the following definition. DEFINITION OF THE NUMBER e Geometrically, this means that of all the possible exponential functions , the function is the one whose tangent line at ( has a slope that is exactly 1. (See Figures 6 and 7.) If we put and, therefore, in Equation 4, it becomes the following impor- tant differentiation formula. fH11032H208490H20850 H33527 1a H33527 e FIGURE 7 0 y 1 x slope=1 slope=e® y=e® {x, e®} 0 y 1 x y=2® y=e® y=3® FIGURE 6 fH11032H208490H208500, 1H20850f H20849xH20850 H33527 e x y H33527 a x lim hl0 e h H11002 1 h H33527 1e is the number such that e fH11032H208490H20850 H33527 1a a H33527 3a H33527 2fH11032H208490H20850fH11032H208490H20850 H33527 1 a d dx H208493 x H20850 H11015 H208491.10H208503 x d dx H208492 x H20850 H11015 H208490.69H208502 x 5 d dx H208493 x H20850 H20895 xH335270 H11015 1.098612 d dx H208492 x H20850 H20895 xH335270 H11015 0.693147 fH11032H208490H20850 H33527 lim hl0 3 h H11002 1 h H11015 1.10for a H33527 3, fH11032H208490H20850 H33527 lim hl0 2 h H11002 1 h H11015 0.69for a H33527 2, a H33527 3a H33527 2 fH11032H208490H20850 SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS |||| 179 h 0.1 0.7177 1.1612 0.01 0.6956 1.1047 0.001 0.6934 1.0992 0.0001 0.6932 1.0987 3 h H11002 1 h 2 h H11002 1 h N In Exercise 1 we will see that lies between and . Later we will be able to show that, correct to five decimal places, e H11015 2.71828 2.82.7 e DERIVATIVE OF THE NATURAL EXPONENTIAL FUNCTION Thus the exponential function has the property that it is its own derivative. The geometrical significance of this fact is that the slope of a tangent line to the curve is equal to the -coordinate of the point (see Figure 7). EXAMPLE 8 If , find and . Compare the graphs of and . SOLUTION Using the Difference Rule, we have In Section 2.8 we defined the second derivative as the derivative of , so The function f and its derivative are graphed in Figure 8. Notice that has a horizon- tal tangent when ; this corresponds to the fact that . Notice also that, for , is positive and is increasing. When , is negative and is decreasing. M EXAMPLE 9 At what point on the curve is the tangent line parallel to the line ? SOLUTION Since , we have . Let the x-coordinate of the point in question be a. Then the slope of the tangent line at that point is . This tangent line will be parallel to the line if it has the same slope, that is, 2. Equating slopes, we get Therefore the required point is . (See Figure 9.) MH20849a, e a H20850 H33527 H20849ln 2, 2H20850 a H33527 ln 2e a H33527 2 y H33527 2x e a yH11032 H33527 e x y H33527 e x y H33527 2x y H33527 e x ffH11032H20849xH20850x H11021 0ffH11032H20849xH20850x H11022 0 fH11032H208490H20850 H33527 0x H33527 0 ffH11032 f H11033H20849xH20850 H33527 d dx H20849e x H11002 1H20850 H33527 d dx H20849e x H20850 H11002 d dx H208491H20850 H33527 e x fH11032 fH11032H20849xH20850 H33527 d dx H20849e x H11002 xH20850 H33527 d dx H20849e x H20850 H11002 d dx H20849xH20850 H33527 e x H11002 1 fH11032ff H11033fH11032f H20849xH20850 H33527 e x H11002 xV yy H33527 e x f H20849xH20850 H33527 e x d dx H20849e x H20850 H33527 e x FIGURE 8 3 _1 1.5_1.5 f fª Visual 3.1 uses the slope-a-scope to illustrate this formula. TEC (b) What types of functions are and ? Compare the differentiation formulas for and t. (c) Which of the two functions in part (b) grows more rapidly when x is large? 3–32 Differentiate the function. 3. 4. 5. 6. 7. 8. f H20849tH20850 H33527 1 2 t 6 H11002 3t 4 H11001 tf H20849xH20850 H33527 x 3 H11002 4x H11001 6 FH20849xH20850 H33527 3 4 x 8 f H20849tH20850 H33527 2 H11002 2 3 t f H20849xH20850 H33527s30 f H20849xH20850 H33527 186.5 f tH20849xH20850 H33527 x e f H20849xH20850 H33527 e x 1. (a) How is the number e defined? (b) Use a calculator to estimate the values of the limits and correct to two decimal places. What can you conclude about the value of e? 2. (a) Sketch, by hand, the graph of the function , pay- ing particular attention to how the graph crosses the y-axis. What fact allows you to do this? f H20849xH20850 H33527 e x lim h l 0 2.8 h H11002 1 h lim h l 0 2.7 h H11002 1 h EXERCISES3.1 FIGURE 9 1 1 0 x 2 3 y y=´ y=2x (ln 2, 2) 180 |||| CHAPTER 3 DIFFERENTIATION RULES (b) Using the graph in part (a) to estimate slopes, make a rough sketch, by hand, of the graph of . (See Example 1 in Section 2.8.) (c) Calculate and use this expression, with a graphing device, to graph . Compare with your sketch in part (b). ; 44. (a) Use a graphing calculator or computer to graph the func- tion in the viewing rectangle by . (b) Using the graph in part (a) to estimate slopes, make a rough sketch, by hand, of the graph of . (See Example 1 in Section 2.8.) (c) Calculate and use this expression, with a graphing device, to graph . Compare with your sketch in part (b). 45–46 Find the first and second derivatives of the function. 45. ; 47–48 Find the first and second derivatives of the function. Check to see that your answers are reasonable by comparing the graphs of , , and . 47. 48. The equation of motion of a particle is , where is in meters and is in seconds. Find (a) the velocity and acceleration as functions of , (b) the acceleration after 2 s, and (c) the acceleration when the velocity is 0. 50. The equation of motion of a particle is , where is in meters and is in seconds. (a) Find the velocity and acceleration as functions of . (b) Find the acceleration after 1 s. ; (c) Graph the position, velocity, and acceleration functions on the same screen. Find the points on the curve where the tangent is horizontal. 52. For what values of does the graph of have a horizontal tangent? 53. Show that the curve has no tangent line with slope 4. 54. Find an equation of the tangent line to the curve that is parallel to the line . 55. Find equations of both lines that are tangent to the curve and are parallel to the line . ; 56. At what point on the curve is the tangent line parallel to the line ? Illustrate by graphing the curve and both lines. 57. Find an equation of the normal line to the parabola that is parallel to the line .x H11002 3y H33527 5y H33527 x 2 H11002 5x H11001 4 3x H11002 y H33527 5 y H33527 1 H11001 2e x H11002 3x 12x H11002 y H33527 1y H33527 1 H11001 x 3 y H33527 1 H11001 3x y H33527 xsx y H33527 6x 3 H11001 5x H11002 3 f H20849xH20850 H33527 x 3 H11001 3x 2 H11001 x H11001 3 x y H33527 2x 3 H11001 3x 2 H11002 12x H11001 151. t tss H33527 2t 3 H11002 7t 2 H11001 4t H11001 1 t t ss H33527 t 3 H11002 3t49. f H20849xH20850 H33527 e x H11002 x 3 f H20849xH20850 H33527 2x H11002 5x 3H208624 f H11033fH11032f GH20849rH20850 H33527sr H11001s 3 r 46.f H20849xH20850 H33527 x 4 H11002 3x 3 H11001 16x tH11032 tH11032H20849xH20850 tH11032 H20851H110028, 8H20852 H20851H110021, 4H20852tH20849xH20850 H33527 e x H11002 3x 2 fH11032 fH11032H20849xH20850 fH11032 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 24. 25. 26. 27. 28. 29. 30. 32. 33–34 Find an equation of the tangent line to the curve at the given point. 33. , 34. , 35–36 Find equations of the tangent line and normal line to the curve at the given point. , 36. , ; 37–38 Find an equation of the tangent line to the curve at the given point. Illustrate by graphing the curve and the tangent line on the same screen. 37. , 38. , ; 39–42 Find . Compare the graphs of and and use them to explain why your answer is reasonable. 39. 40. 41. 42. ; 43. (a) Use a graphing calculator or computer to graph the func- tion in the viewing rectangle by .H20851H1100210, 50H20852H20851H110023, 5H20852 f H20849xH20850 H33527 x 4 H11002 3x 3 H11002 6x 2 H11001 7x H11001 30 f H20849xH20850 H33527 x H11001 1 x f H20849xH20850 H33527 3x 15 H11002 5x 3 H11001 3 f H20849xH20850 H33527 3x 5 H11002 20x 3 H11001 50xf H20849xH20850 H33527 e x H11002 5x fH11032ffH11032H20849xH20850 H208491, 0H20850y H33527 x H11002sx H208491, 2H20850y H33527 3x 2 H11002 x 3 H208491, 9H20850y H33527 H208491 H11001 2xH20850 2 H208490, 2H20850y H33527 x 4 H11001 2e x 35. H208491, 2H20850y H33527 x 4 H11001 2x 2 H11002 xH208491, 1H20850y H33527s 4 x y H33527 e xH110011 H11001 1z H33527 A y 10 H11001 Be y 31. v H33527 H20873 sx H11001 1 s 3 x H20874 2 u H33527s 5 t H11001 4st 5 y H33527 ae v H11001 b v H11001 c v 2 HH20849xH20850 H33527 H20849x H11001 x H110021 H20850 3 tH20849uH20850 H33527s2 u H11001s3u y H33527 4H9266 2 y H33527 x 2 H11002 2sx x y H33527 x 2 H11001 4x H11001 3 sx 23. y H33527sx H20849x H11002 1H20850y H33527 ax 2 H11001 bx H11001 c f H20849tH20850 H33527st H11002 1 st F H20849xH20850 H33527 ( 1 2 x) 5 y H33527s 3 x GH20849xH20850 H33527sx H11002 2e x BH20849yH20850 H33527 cy H110026 AH20849sH20850 H33527 H11002 12 s 5 RH20849tH20850 H33527 5t H110023H208625 VH20849rH20850 H33527 4 3 H9266r 3 y H33527 5e x H11001 3y H33527 x H110022H208625 hH20849xH20850 H33527 H20849x H11002 2H20850H208492x H11001 3H20850f H20849tH20850 H33527 1 4 H20849t 4 H11001 8H20850 SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS |||| 181 69. (a) For what values of is the function dif- ferentiable? Find a formula for . (b) Sketch the graphs of and . 70. Where is the function differenti- able? Give a formula for and sketch the graphs of and . 71. Find the parabola with equation whose tangent line at (1, 1) has equation . 72. Suppose the curve has a tan- gent line when with equation and a tangent line when with equation . Find the values of , , , and . For what values of and is the line tangent to the parabola when ? 74. Find the value of such that the line is tangent to the curve . 75. Let Find the values of and that make differentiable every- where. 76. A tangent line is drawn to the hyperbola at a point . (a) Show that the midpoint of the line segment cut from this tangent line by the coordinate axes is . (b) Show that the triangle formed by the tangent line and the coordinate axes always has the same area, no matter where is located on the hyperbola. Evaluate . 78. Draw a diagram showing two perpendicular lines that intersect on the -axis and are both tangent to the parabola . Where do these lines intersect? 79. If , how many lines through the point are normal lines to the parabola ? What if ? 80. Sketch the parabolas and . Do you think there is a line that is tangent to both curves? If so, find its equation. If not, why not? y H33527 x 2 H11002 2x H11001 2y H33527 x 2 c H33355 1 2 y H33527 x 2 H208490, cH20850c H11022 1 2 y H33527 x 2 y lim x l 1 x 1000 H11002 1 x H11002 1 77. P P Pxy H33527 c fbm f H20849xH20850 H33527 H20877 x 2 mx H11001 b if x H33355 2 if x H11022 2 y H33527 csx y H33527 3 2 x H11001 6c x H33527 2y H33527 ax 2 2x H11001 y H33527 bba 73. dcba y H33527 2 H11002 3xx H33527 1 y H33527 2x H11001 1x H33527 0 y H33527 x 4 H11001 ax 3 H11001 bx 2 H11001 cx H11001 d y H33527 3x H11002 2 y H33527 ax 2 H11001 bx hH11032hhH11032 hH20849xH20850 H33527 H11341 x H11002 1 H11341 H11001 H11341 x H11001 2 H11341 fH11032f fH11032 f H20849xH20850 H33527 H11341 x 2 H11002 9 H11341 x 58. Where does the normal line to the parabola at the point (1, 0) intersect the parabola a second time? Illustrate with a sketch. Draw a diagram to show that there are two tangent lines to the parabola that pass through the point . Find the coordinates of the points where these tangent lines inter- sect the parabola. 60. (a) Find equations of both lines through the point that are tangent to the parabola . (b) Show that there is no line through the point that is tangent to the parabola. Then draw a diagram to see why. 61. Use the definition of a derivative to show that if , then . (This proves the Power Rule for the case .) 62. Find the derivative of each function by calculating the first few derivatives and observing the pattern that occurs. (a) (b) 63. Find a second-degree polynomial such that , , and . 64. The equation is called a differential equation because it involves an unknown function and its derivatives and . Find constants such that the function satisfies this equation. (Differ- ential equations will be studied in detail in Chapter 9.) 65. Find a cubic function whose graph has horizontal tangents at the points and . 66. Find a parabola with equation that has slope 4 at , slope at , and passes through the point . 67. Let Is differentiable at 1? Sketch the graphs of and . 68. At what numbers is the following function differentiable? Give a formula for and sketch the graphs of and .tH11032ttH11032 tH20849xH20850 H33527 H20877 H110021 H11002 2x x 2 x if x H11021 H110021 if H110021 H33355 x H33355 1 if x H11022 1 t fH11032ff f H20849xH20850 H33527 H20877 2 H11002 x x 2 H11002 2x H11001 2 if x H33355 1 if x H11022 1 H208492, 15H20850 x H33527 H110021H110028x H33527 1 y H33527 ax 2 H11001 bx H11001 c H208492, 0H20850H20849H110022, 6H20850 y H33527 ax 3 H11001 bx 2 H11001 cx H11001 d y H33527 Ax 2 H11001 Bx H11001 C A, B, and CyH11033yH11032 y yH11033 H11001 yH11032 H11002 2y H33527 x 2 PH11033H208492H20850 H33527 2PH11032H208492H20850 H33527 3 PH208492H20850 H33527 5P f H20849xH20850 H33527 1H20862xf H20849xH20850 H33527 x n nth n H33527 H110021 fH11032H20849xH20850 H33527 H110021H20862x 2 f H20849xH20850 H33527 1H20862x H208492, 7H20850 y H33527 x 2 H11001 x H208492, H110023H20850 H208490, H110024H20850y H33527 x 2 59. y H33527 x H11002 x 2 182 |||| CHAPTER 3 DIFFERENTIATION RULES Suppose you are asked to design the first ascent and drop for a new roller coaster. By studying photographs of your favorite coasters, you decide to make the slope of the ascent 0.8 and the slope of the drop . You decide to connect these two straight stretches and with part of a parabola , where and are measured in feet. For the track to be smooth there can’t be abrupt changes in direction, so you want the linear f H20849xH20850xy H33527 f H20849xH20850 H33527 ax 2 H11001 bx H11001 cy H33527 L 2 H20849xH20850 y H33527 L 1 H20849xH20850H110021.6 BUILDING A BETTER ROLLER COASTERA P P L I E D P R O J E C T James Stewart Stewart - Calculus - Early Transcedentals 6e calculus