Differentiation Rules 159

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3.2 Differentiation Rules

This section introduces a few rules that allow us to differentiate a great variety of functions. By proving these rules here, we can differentiate functions without having to apply the definition of the derivative each time.

Powers, Multiples, Sums, and Differences The first rule of differentiation is that the derivative of every constant function is zero.

RULE 1 Derivative of a Constant Function If ƒ has the constant value ƒsxd = c, then dƒ d = scd = 0. dx dx

EXAMPLE 1

y If ƒ has the constant value ƒsxd = 8, then

c (x, c) (x h, c) df d = s8d = 0. y c dx dx Similarly,

h d p d x - = 0 and 23 = 0. 0 x x h dx 2 dx a b a b

FIGURE 3.8 The rule sd dxdscd = 0 is Proof of Rule 1 We apply the definition of derivative to ƒsxd = c, the function whose > another way to say that the values of outputs have the constant value c (Figure 3.8). At every value of x, we find that constant functions never change and that the slope of a horizontal line is zero at ƒsx + hd - ƒsxd c - c ƒ¿sxd = lim = lim = lim 0 = 0. every point. h:0 h h:0 h h:0 4100 AWL/Thomas_ch03p147-243 8/19/04 11:16 AM Page 160

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The second rule tells how to differentiate xn if n is a positive integer.

RULE 2 Power Rule for Positive Integers If n is a positive integer, then

d - xn = nxn 1 . dx

To apply the Power Rule, we subtract 1 from the original exponent (n) and multiply the result by n.

EXAMPLE 2 Interpreting Rule 2 ƒ x x2 x3 x4 Á ƒ¿ 12x 3x2 4x3 Á

HISTORICAL BIOGRAPHY First Proof of Rule 2 The formula Richard Courant n n n-1 n-2 Á n-2 n-1 (1888–1972) z - x = sz - xdsz + z x + + zx + x d can be verified by multiplying out the right-hand side. Then from the alternative form for the definition of the derivative, ƒszd - ƒsxd zn - xn ƒ ¿sxd = lim - = lim - z:x z x z:x z x = limszn-1 + zn-2x + Á + zxn-2 + xn-1d z:x = nxn-1

Second Proof of Rule 2 If ƒsxd = xn , then ƒsx + hd = sx + hdn . Since n is a positive integer, we can expand sx + hdn by the Binomial Theorem to get

ƒsx + hd - ƒsxd sx + hdn - xn ƒ ¿sxd = lim = lim h:0 h h:0 h

- nsn - 1d - - xn + nxn 1h + xn 2h2 + Á + nxhn 1 + hn - xn c 2 d = lim h:0 h

- nsn - 1d - - nxn 1h + xn 2h2 + Á + nxhn 1 + hn 2 = lim h:0 h

- nsn - 1d - - - = lim nxn 1 + xn 2h + Á + nxhn 2 + hn 1 h:0 c 2 d = nxn-1 The third rule says that when a differentiable function is multiplied by a constant, its derivative is multiplied by the same constant. 4100 AWL/Thomas_ch03p147-243 8/19/04 11:16 AM Page 161

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RULE 3 Constant Multiple Rule If u is a differentiable function of x, and c is a constant, then d du scud = c . dx dx

y In particular, if n is a positive integer, then y 3x2 d - scxnd = cnxn 1 . dx

Slope 3(2x) EXAMPLE 3 6x (a) The derivative formula 3 (1, 3) Slope 6(1) 6

d 2 # 2 s3x d = 3 2x = 6x y x dx 2 says that if we rescale the graph of y = x2 by multiplying each y-coordinate by 3, then we multiply the slope at each point by 3 (Figure 3.9).

Slope 2x (b) A useful special case 1 Slope 2(1) 2 (1, 1) The derivative of the negative of a differentiable function u is the negative of the function’s derivative. Rule 3 with c =-1 gives

d - = d - # =- # d =-du x s ud s 1 ud 1 sud . 0 1 2 dx dx dx dx

FIGURE 3.9 The graphs of y = x2 and Proof of Rule 3 y = 3x2 . Tripling the y-coordinates triples d cusx + hd - cusxd Derivative definition the slope (Example 3). cu = lim dx h:0 h with ƒsxd = cusxd usx + hd - usxd = c lim Limit property h:0 h du = c u is differentiable. dx The next rule says that the derivative of the sum of two differentiable functions is the sum of their derivatives. Denoting Functions by u and Y The functions we are working with when we need a differentiation formula are likely to be denoted by letters like ƒ RULE 4 Derivative Sum Rule and g. When we apply the formula, we + do not want to find it using these same If u and y are differentiable functions of x, then their sum u y is differentiable letters in some other way. To guard at every point where u and y are both differentiable. At such points, against this problem, we denote the d du dy functions in differentiation rules by su + yd = + . letters like u and y that are not likely to dx dx dx be already in use. 4100 AWL/Thomas_ch03p147-243 8/19/04 11:16 AM Page 162

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EXAMPLE 4 Derivative of a Sum y = x4 + 12x dy d d = sx4d + s12xd dx dx dx = 4x3 + 12

Proof of Rule 4 We apply the definition of derivative to ƒsxd = usxd + ysxd:

d [usx + hd + ysx + hd] - [usxd + ysxd] [usxd + ysxd] = lim dx h:0 h usx + hd - usxd ysx + hd - ysxd = lim + h:0 c h h d usx + hd - usxd ysx + hd - ysxd du dy = lim + lim = + . h:0 h h:0 h dx dx Combining the Sum Rule with the Constant Multiple Rule gives the Difference Rule, which says that the derivative of a difference of differentiable functions is the difference of their derivatives. d d du dy du dy su - yd = [u + s -1dy] = + s -1d = - dx dx dx dx dx dx The Sum Rule also extends to sums of more than two functions, as long as there are only finitely many functions in the sum. If u1 , u2 , Á , un are differentiable at x, then so is Á u1 + u2 + + un , and

d du1 du2 dun su + u + Á + u d = + + Á + . dx 1 2 n dx dx dx

EXAMPLE 5 Derivative of a Polynomial 4 y = x3 + x2 - 5x + 1 3 dy d d 4 d d = x3 + x2 - s5xd + s1d dx dx dx a3 b dx dx 4 = 3x2 + # 2x - 5 + 0 3 8 = 3x2 + x - 5 3 Notice that we can differentiate any polynomial term by term, the way we differentiated the polynomial in Example 5. All polynomials are differentiable everywhere.

Proof of the Sum Rule for Sums of More Than Two Functions We prove the statement

d du1 du2 dun su + u + Á + u d = + + Á + dx 1 2 n dx dx dx by mathematical induction (see Appendix 1). The statement is true for n = 2, as was just proved. This is Step 1 of the induction proof. 4100 AWL/Thomas_ch03p147-243 8/20/04 9:12 AM Page 163

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Step 2 is to show that if the statement is true for any positive integer n = k, where k Ú n0 = 2, then it is also true for n = k + 1. So suppose that

d du1 du2 duk su + u + Á + u d = + + Á + . (1) dx 1 2 k dx dx dx Then d Á (u + u + + u + u + ) dx (++++)++++*1 2 k ()*k 1 Call the function Call this defined by this sum u. function y.

d Á duk+1 d = su + u + + u d + Rule 4 for su + yd dx 1 2 k dx dx

du1 du2 Á duk duk+1 = + + + + . Eq. (1) dx dx dx dx With these steps verified, the mathematical induction principle now guarantees the Sum Rule for every integer n Ú 2. EXAMPLE 6 Finding Horizontal Tangents 4 2 y y ϭ x4 Ϫ 2x2 ϩ 2 Does the curve y = x - 2x + 2 have any horizontal tangents? If so, where?

Solution The horizontal tangents, if any, occur where the slope dy dx is zero. We have, > dy d (0, 2) = sx4 - 2x2 + 2d = 4x3 - 4x. dx dx dy 1 Now solve the equation = 0 for x: (–1, 1) (1, 1) dx 4 x3 - 4x = 0 x –1 0 1 4 xsx2 - 1d = 0 x = 0, 1, -1. FIGURE 3.10 The curve y = x4 - 2x2 + 2 and its horizontal The curve y = x4 - 2x2 + 2 has horizontal tangents at x = 0, 1, and -1. The corre- tangents (Example 6). sponding points on the curve are (0, 2), (1, 1) and s -1, 1d. See Figure 3.10.

Products and Quotients While the derivative of the sum of two functions is the sum of their derivatives, the derivative of the product of two functions is not the product of their derivatives. For instance, d d d d sx # xd = sx2d = 2x, while sxd # sxd = 1 # 1 = 1. dx dx dx dx The derivative of a product of two functions is the sum of two products, as we now explain.

RULE 5 Derivative Product Rule If u and y are differentiable at x, then so is their product uy, and d dy du suyd = u + y . dx dx dx 4100 AWL/Thomas_ch03p147-243 8/19/04 11:16 AM Page 164

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y y y Picturing the Product Rule The derivative of the product u is u times the derivative of plus times the deriva- s yd¿= y¿+y ¿ If u(x) and y(x) are positive and tive of u. In prime notation, u u u . In function notation, increase when x increases, and if h 7 0, d [ƒsxdgsxd] = ƒsxdg¿sxd + gsxdƒ¿sxd. dx y(x h) y u(x) y u y EXAMPLE 7 Using the Product Rule y(x) Find the derivative of u(x)y(x) y(x) u 1 1 y = x2 + . x a x b 0 u u(x) u(x h) Solution We apply the Product Rule with u = 1 x and y = x2 + s1 xd: > > d dy du then the total shaded area in the picture d 1 1 1 1 1 1 suyd = u + y , and 2 + = - + 2 + - dx dx dx x x x x 2x 2 x x 2 is dx c a bd a x b a ba x b d 1 1 =- by usx + hdysx + hd - usxdysxd x 2 1 1 dx a b x = usx + hd ¢y + ysx + hd = 2 - - 1 - Example 3, Section 2.7. x3 x3 ¢u - ¢u¢y. 2 Dividing both sides of this equation by = 1 - . h gives x3 usx + hdysx + hd - usxdysxd h Proof of Rule 5 ¢y ¢u = usx + hd + ysx + hd d usx + hdysx + hd - usxdysxd h h suyd = lim dx : h ¢y h 0 - ¢u . h To change this fraction into an equivalent one that contains difference quotients for the de- As h : 0+ , rivatives of u and y, we subtract and add usx + hdysxd in the numerator: ¢y dy ¢u # : 0 # = 0, d usx + hdysx + hd - usx + hdysxd + usx + hdysxd - usxdysxd h dx suyd = lim dx : h leaving h 0 + - + - d dy du ysx hd ysxd usx hd usxd suyd = u + y . = lim usx + hd + ysxd dx dx dx h:0 c h h d ysx + hd - ysxd usx + hd - usxd = lim usx + hd # lim + ysxd # lim . h:0 h:0 h h:0 h As h approaches zero,usx + hd approaches u(x) because u, being differentiable at x, is continuous at x. The two fractions approach the values of dy dx at x and du dx at x. In short, > > d dy du suyd = u + y . dx dx dx In the following example, we have only numerical values with which to work.

EXAMPLE 8 Derivative from Numerical Values Let y = uy be the product of the functions u and y. Find y¿s2d if us2d = 3, u¿s2d =-4, ys2d = 1, and y¿s2d = 2. Solution From the Product Rule, in the form y¿=suyd¿=uy¿+yu¿ , 4100 AWL/Thomas_ch03p147-243 8/19/04 11:16 AM Page 165

3.2 Differentiation Rules 165

we have y¿s2d = us2dy¿s2d + ys2du¿s2d = s3ds2d + s1ds -4d = 6 - 4 = 2.

EXAMPLE 9 Differentiating a Product in Two Ways Find the derivative of y = sx2 + 1dsx3 + 3d.

Solution (a) From the Product Rule with u = x2 + 1 and y = x3 + 3, we find d x2 + 1 x3 + 3 = sx2 + 1ds3x2d + sx3 + 3ds2xd dx CA BA BD = 3x4 + 3x2 + 2x4 + 6x = 5x4 + 3x2 + 6x. (b) This particular product can be differentiated as well (perhaps better) by multiplying out the original expression for y and differentiating the resulting polynomial: y = sx2 + 1dsx3 + 3d = x5 + x3 + 3x2 + 3 dy = 5x4 + 3x2 + 6x. dx This is in agreement with our first calculation. Just as the derivative of the product of two differentiable functions is not the product of their derivatives, the derivative of the quotient of two functions is not the quotient of their derivatives. What happens instead is the Quotient Rule.

RULE 6 Derivative Quotient Rule If u and y are differentiable at x and if ysxd Z 0, then the quotient u y is differ- > entiable at x, and

du dy y - u d u dx dx = . dx ay b y2

In function notation, d ƒsxd gsxdƒ¿sxd - ƒsxdg¿sxd = . dx c gsxd d g2sxd

EXAMPLE 10 Using the Quotient Rule Find the derivative of t 2 - 1 y = . t 2 + 1 4100 AWL/Thomas_ch03p147-243 8/19/04 11:16 AM Page 166

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Solution We apply the Quotient Rule with u = t 2 - 1 and y = t 2 + 1: 2 # 2 # dy st + 1d 2t - st - 1d 2t d u ysdu dtd - usdy dtd = = > > dt st 2 + 1d2 dt ay b y2 2t 3 + 2t - 2t 3 + 2t = st 2 + 1d2 4t = . st 2 + 1d2

Proof of Rule 6 usx + hd usxd - d u ysx + hd ysxd = lim dx ay b h:0 h ysxdusx + hd - usxdysx + hd = lim h:0 hysx + hdysxd To change the last fraction into an equivalent one that contains the difference quotients for the derivatives of u and y, we subtract and add y(x)u(x) in the numerator. We then get

d u ysxdusx + hd - ysxdusxd + ysxdusxd - usxdysx + hd = lim dx ay b h:0 hysx + hdysxd usx + hd - usxd ysx + hd - ysxd ysxd - usxd h h = lim . h:0 ysx + hdysxd Taking the limit in the numerator and denominator now gives the Quotient Rule.

Negative Integer Powers of x The Power Rule for negative integers is the same as the rule for positive integers.

RULE 7 Power Rule for Negative Integers If n is a negative integer and x Z 0, then

d - sxnd = nxn 1 . dx

EXAMPLE 11

d 1 d - - 1 (a) = sx 1d = s -1dx 2 =- Agrees with Example 3, Section 2.7 dx ax b dx x2

d 4 d - - 12 (b) = 4 sx 3d = 4s -3dx 4 =- dx ax3 b dx x4 4100 AWL/Thomas_ch03p147-243 8/19/04 11:16 AM Page 167

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Proof of Rule 7 The proof uses the Quotient Rule. If n is a negative integer, then n =-m, where m is a positive integer. Hence, xn = x -m = 1 xm , and > d d 1 sxnd = dx dx axm b d d xm # 1 - 1 # xm y dx dx = A B A B Quotient Rule with u = 1 and y = xm sxmd2 2 - m-1 y x 0 mx d m m-1 x = Since m 7 0, sx d = mx x2m dx =- -m-1 4 mx - = nxn 1 . Since -m = n 3 (1, 3)

2 EXAMPLE 12 Tangent to a Curve Find an equation for the tangent to the curve 1 y –x 4 = + 2 y x x x 0 1 2 3 at the point (1, 3) (Figure 3.11). FIGURE 3.11 The tangent to the curve y = x + s2 xd at (1, 3) in Example 12. > Solution The slope of the curve is The curve has a third-quadrant portion dy not shown here. We see how to graph = d + d 1 = + - 1 = - 2 sxd 2 x 1 2 2 1 2 . functions like this one in Chapter 4. dx dx dx a b a x b x The slope at x = 1 is dy = - 2 = - =- 1 2 1 2 1. dx ` x=1 c x d x=1 The line through (1, 3) with slope m =-1 is

y - 3 = s -1dsx - 1d Point-slope equation y =-x + 1 + 3 y =-x + 4. The choice of which rules to use in solving a differentiation problem can make a difference in how much work you have to do. Here is an example.

EXAMPLE 13 Choosing Which Rule to Use Rather than using the Quotient Rule to find the derivative of sx - 1dsx2 - 2xd y = , x4 expand the numerator and divide by x4 :

2 3 2 sx - 1dsx - 2xd x - 3x + 2x - - - y = = = x 1 - 3x 2 + 2x 3 . x4 x4 4100 AWL/Thomas_ch03p147-243 8/19/04 11:16 AM Page 168

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Then use the Sum and Power Rules:

dy =-x -2 - 3s -2dx -3 + 2s -3dx -4 dx 1 6 6 =- + - . x2 x3 x4

Second- and Higher-Order Derivatives If y = ƒsxd is a differentiable function, then its derivative ƒ¿sxd is also a function. If ƒ¿ is also differentiable, then we can differentiate ƒ¿ to get a new function of x denoted by ƒ– . So ƒ–=sƒ¿d¿ . The function ƒ– is called the second derivative of ƒ because it is the derivative of the first derivative. Notationally, 2 ¿ d y d dy dy 2 2 ƒ–sxd = = = = y–=D sƒdsxd = D x ƒsxd. dx2 dx adx b dx

The symbol D 2 means the operation of differentiation is performed twice. If y = x6 , then y¿=6x5 and we have

dy¿ d y–= = 6x5 = 30x4 . dx dx A B Thus D 2 x6 = 30x4 . A B If y– is differentiable, its derivative, y‡=dy– dx = d 3y dx3 is the third derivative > > How to Read the Symbols for of y with respect to x. The names continue as you imagine, with Derivatives n y¿ “y prime” d - d y y snd = y sn 1d = = D ny y– “y double prime” dx dxn d 2y “d squared y dx squared” dx2 denoting the nth derivative of y with respect to x for any positive integer n. y‡ “y triple prime” We can interpret the second derivative as the rate of change of the slope of the tangent y snd “y super n” to the graph of y = ƒsxd at each point. You will see in the next chapter that the second de- d ny rivative reveals whether the graph bends upward or downward from the tangent line as we “d to the n of y by dx to the n” dxn move off the point of tangency. In the next section, we interpret both the second and third D n “D to the n” derivatives in terms of motion along a straight line.

EXAMPLE 14 Finding Higher Derivatives The first four derivatives of y = x3 - 3x2 + 2 are

First derivative: y¿=3x2 - 6x Second derivative: y–=6x - 6 Third derivative: y‡=6 Fourth derivative: y s4d = 0.

The function has derivatives of all orders, the fifth and later derivatives all being zero. 4100 AWL/Thomas_ch03p147-243 8/19/04 11:16 AM Page 169

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EXERCISES 3.2

Derivative Calculations q2 + 3 q4 - 1 q2 + 3 37.p = 38. p = In Exercises 1–12, find the first and second derivatives. a 12q ba q3 b sq - 1d3 + sq + 1d3 1.y =-x2 + 3 2. y = x2 + x + 8 3.s = 5t 3 - 3t 5 4. w = 3z7 - 7z3 + 21z2 Using Numerical Values 4x3 x3 x2 x 39. Suppose u and y are functions of x that are differentiable at x = 0 5.y = - x 6. y = + + 3 3 2 4 and that

- 1 - 4 us0d = 5, u¿s0d =-3, ys0d =-1, y¿s0d = 2. 7.w = 3z 2 - 8. s =-2t 1 + z t 2 Find the values of the following derivatives at x = 0. 9.y = 6x2 - 10x - 5x -2 10. y = 4 - 2x - x -3 d d u d y d - 1 5 12 4 1 a. suyd b. y c. u d. s7y 2ud 11.r = - 12. r = - + dx dx a b dx a b dx 3s2 2s u u3 u4 40. Suppose u and y are differentiable functions of x and that In Exercises 13–16, find y¿ (a) by applying the Product Rule and us1d = 2, u¿s1d = 0, ys1d = 5, y¿s1d =-1. (b) by multiplying the factors to produce a sum of simpler terms to differentiate. Find the values of the following derivatives at x = 1. 13.y = s3 - x2dsx3 - x + 1d 14. y = sx - 1dsx2 + x + 1d d d u d y d - a. suyd b. y c. u d. s7y 2ud 1 1 1 dx dx a b dx a b dx 15.y = sx2 + 1d x + 5 + 16. y = x + x - + 1 a x b a x ba x b

Find the derivatives of the functions in Exercises 17–28. Slopes and Tangents 2x + 5 2x + 1 41. a. Normal to a curve Find an equation for the line perpendicular 17.y = 18. z = to the tangent to the curve y = x3 - 4x + 1 at the point (2, 1). 3x - 2 x2 - 1 b. Smallest slope What is the smallest slope on the curve? At x2 - 4 t 2 - 1 19.gsxd = 20. ƒstd = what point on the curve does the curve have this slope? x + 0.5 t 2 + t - 2 c. Tangents having specified slope Find equations for the 21.y = s1 - tds1 + t 2d-1 22. w = s2x - 7d-1sx + 5d tangents to the curve at the points where the slope of the 1s - 1 5x + 1 23.ƒssd = 24. u = curve is 8. + 1s 1 21x 42. a. Horizontal tangents Find equations for the horizontal tan- + - 1 = 3 - - = 1 x 4 x = 1 + gents to the curve y x 3x 2. Also find equations for 25.y x 26. r 2 2u a 2u b the lines that are perpendicular to these tangents at the points of tangency. 1 sx + 1dsx + 2d 27.y = 28. y = sx2 - 1dsx2 + x + 1d sx - 1dsx - 2d b. Smallest slope What is the smallest slope on the curve? At what point on the curve does the curve have this slope? Find Find the derivatives of all orders of the functions in Exercises 29 and an equation for the line that is perpendicular to the curve’s 30. tangent at this point. x4 3 x5 43. Find the tangents to Newton’s serpentine (graphed here) at the ori- 29.y = - x2 - x 30. y = 2 2 120 gin and the point (1, 2). Find the first and second derivatives of the functions in Exercises y 31–38. y 4x 3 + 2 + - 2 = x 7 = t 5t 1 x 1 31.y x 32. s 2 (1, 2) t 2 su - 1dsu2 + u + 1d sx2 + xdsx2 - x + 1d 1 33.r = 34. u = 3 4 x u x 0 1 2 3 4 1 + 3z 35.w = s3 - zd 36. w = sz + 1dsz - 1dsz2 + 1d a 3z b 4100 AWL/Thomas_ch03p147-243 8/19/04 11:16 AM Page 170

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44. Find the tangent to the Witch of Agnesi (graphed here) at the point how to find the amount of medicine to which the body is most (2, 1). sensitive.

y 51. Suppose that the function y in the Product Rule has a constant y 8 value c. What does the Product Rule then say? What does this say x2 4 about the Constant Multiple Rule? 2 52. The Reciprocal Rule (2, 1) 1 a. The Reciprocal Rule says that at any point where the function x 0 1 2 3 y(x) is differentiable and different from zero,

45. Quadratic tangent to identity function The curve y = d 1 =-1 dy y 2 . ax2 + bx + c passes through the point (1, 2) and is tangent to the dx a b y dx line y = x at the origin. Find a, b, and c. Show that the Reciprocal Rule is a special case of the 46. Quadratics having a common tangent The curves y = Quotient Rule. x2 + ax + b and y = cx - x2 have a common tangent line at b. Show that the Reciprocal Rule and the Product Rule together the point (1, 0). Find a, b, and c. imply the Quotient Rule. 47. a. Find an equation for the line that is tangent to the curve 53. Generalizing the Product Rule The Product Rule gives the y = x3 - x at the point s -1, 0d. formula b. Graph the curve and tangent line together. The tangent T d dy du suyd = u + y intersects the curve at another point. Use Zoom and Trace to dx dx dx estimate the point’s coordinates. T c. Confirm your estimates of the coordinates of the second for the derivative of the product uy of two differentiable functions intersection point by solving the equations for the curve and of x. tangent simultaneously (Solver key). a. What is the analogous formula for the derivative of the 48. a. Find an equation for the line that is tangent to the curve product uyw of three differentiable functions of x? 3 2 y = x - 6x + 5x at the origin. b. What is the formula for the derivative of the product u1 u2 u3 u4 T b. Graph the curve and tangent together. The tangent intersects of four differentiable functions of x? the curve at another point. Use Zoom and Trace to estimate c. What is the formula for the derivative of a product

the point’s coordinates. u1 u2 u3 Á un of a finite number n of differentiable functions T c. Confirm your estimates of the coordinates of the second of x? intersection point by solving the equations for the curve and 54. Rational Powers tangent simultaneously (Solver key). d 3 2 3 2 # 1 2 a. Find x > by writing x > as x x > and using the Product dx A B Theory and Examples Rule. Express your answer as a rational number times a 49. The general polynomial of degree n has the form rational power of x. Work parts (b) and (c) by a similar method. n n-1 Á 2 Psxd = an x + an-1 x + + a2 x + a1 x + a0 d 5 2 b. Find sx > d. dx where an Z 0. Find P¿sxd.

50. The body’s reaction to medicine The reaction of the body to a d 7 2 c. Find sx > d. dose of medicine can sometimes be represented by an equation of dx the form d. What patterns do you see in your answers to parts (a), (b), and C M (c)? Rational powers are one of the topics in Section 3.6. R = M 2 - , a 2 3 b 55. Cylinder pressure If gas in a cylinder is maintained at a constant temperature T, the pressure P is related to the volume V by a where C is a positive constant and M is the amount of medicine formula of the form absorbed in the blood. If the reaction is a change in blood pres- nRT an2 sure, R is measured in millimeters of mercury. If the reaction is a P = - , - 2 change in temperature, R is measured in degrees, and so on. V nb V Find dR dM . This derivative, as a function of M, is called the in which a, b, n,and R are constants. Find dP dV . (See accompa- > > sensitivity of the body to the medicine. In Section 4.5, we will see nying figure.) 4100 AWL/Thomas_ch03p147-243 8/19/04 11:16 AM Page 171

3.2 Differentiation Rules 171

56. The best quantity to order One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is

km hq Asqd = + cm + , q 2

where q is the quantity you order when things run low (shoes, ra- dios, brooms, or whatever the item might be); k is the cost of plac- ing an order (the same, no matter how often you order); c is the cost of one item (a constant); m is the number of items sold each week (a constant); and h is the weekly holding cost per item (a constant that takes into account things such as space, utilities, in- surance, and security). Find dA dq and d 2A dq2 . > > 4100 AWL/Thomas_ch03p147-243 8/19/04 11:16 AM Page 183

3.4 Derivatives of Trigonometric Functions 183

3.4 Derivatives of Trigonometric Functions

Many of the phenomena we want information about are approximately periodic (electro- magnetic fields, heart rhythms, tides, weather). The derivatives of sines and cosines play a key role in describing periodic changes. This section shows how to differentiate the six basic trigonometric functions.

Derivative of the Sine Function

To calculate the derivative of ƒsxd = sin x, for x measured in radians, we combine the limits in Example 5a and Theorem 7 in Section 2.4 with the angle sum identity for the sine:

sin sx + hd = sin x cos h + cos x sin h. 4100 AWL/Thomas_ch03p147-243 8/19/04 11:16 AM Page 184

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If ƒsxd = sin x, then ƒsx + hd - ƒsxd ƒ ¿sxd = lim h:0 h sin sx + hd - sin x = lim Derivative definition h:0 h ssin x cos h + cos x sin hd - sin x = lim Sine angle sum identity h:0 h sin x scos h - 1d + cos x sin h = lim h:0 h cos h - 1 sin h = lim sin x # + lim cos x # h:0 a h b h:0 a h b cos h - 1 sin h = sin x # lim + cos x # lim h:0 h h:0 h = sin x # 0 + cos x # 1 Example 5(a) and = cos x. Theorem 7, Section 2.4

The derivative of the sine function is the cosine function: d ssin xd = cos x. dx

EXAMPLE 1 Derivatives Involving the Sine

(a) y = x2 - sin x: dy d = 2x - sin x Difference Rule dx dx A B = 2x - cos x. (b) y = x2 sin x: dy d = x2 sin x + 2x sin x Product Rule dx dx A B = x2 cos x + 2x sin x. = sin x (c) y x : d x # sin x - sin x # 1 dy dx = A B Quotient Rule dx x2 x cos x - sin x = . x2

Derivative of the Cosine Function With the help of the angle sum formula for the cosine,

cos sx + hd = cos x cos h - sin x sin h, 4100 AWL/Thomas_ch03p147-243 8/19/04 11:16 AM Page 185

3.4 Derivatives of Trigonometric Functions 185

we have d cossx + hd - cos x scos xd = lim Derivative definition y : dx h 0 h 1 y cos x scos x cos h - sin x sin hd - cos x = Cosine angle sum x lim identity –0 h:0 h – 1 cos xscos h - 1d - sin x sin h y' = lim y' –sin x h:0 h 1 - = # cos h 1 - # sin h x lim cos x lim sin x –0 h:0 h h:0 h –1 cos h - 1 sin h = cos x # lim - sin x # lim h:0 h h:0 h FIGURE 3.23 The curve y¿=-sin x as = # - # the graph of the slopes of the tangents to cos x 0 sin x 1 Example 5(a) and Theorem 7, Section 2.4 the curve y = cos x. =-sin x.

The derivative of the cosine function is the negative of the sine function: d scos xd =-sin x dx

Figure 3.23 shows a way to visualize this result. EXAMPLE 2 Derivatives Involving the Cosine

(a) y = 5x + cos x: dy d d = s5xd + cos x Sum Rule dx dx dx A B = 5 - sin x.

(b) y = sin x cos x: dy d d = sin x cos x + cos x sin x Product Rule dx dx A B dx A B = sin xs -sin xd + cos xscos xd = cos2 x - sin2 x.

cos x (c) y = : 1 - sin x d d 1 - sin x cos x - cos x 1 - sin x dy dx dx = A B A B A B Quotient Rule dx s1 - sin xd2 s1 - sin xds -sin xd - cos xs0 - cos xd = s1 - sin xd2 1 - sin x = sin2 x + cos2 x = 1 s1 - sin xd2 1 = . 1 - sin x 4100 AWL/Thomas_ch03p147-243 8/19/04 11:16 AM Page 186

186 Chapter 3: Differentiation

Simple Harmonic Motion The motion of a body bobbing freely up and down on the end of a spring or bungee cord is an example of simple harmonic motion. The next example describes a case in which there –5 are no opposing forces such as friction or buoyancy to slow the motion down.

EXAMPLE 3 Motion on a Spring Rest 0 position A body hanging from a spring (Figure 3.24) is stretched 5 units beyond its rest position and released at time t = 0 to bob up and down. Its position at any later time t is Position at 5 s = 5 cos t. t 0 s What are its velocity and acceleration at time t ?

FIGURE 3.24 A body hanging from Solution We have avertical spring and then displaced Position: s = 5 cos t oscillates above and below its rest position. ds d Velocity: y = = s5 cos td =-5 sin t Its motion is described by trigonometric dt dt functions (Example 3). dy d Acceleration: a = = s -5 sin td =-5 cos t. dt dt Notice how much we can learn from these equations: s, y 1. As time passes, the weight moves down and up between s =-5 and s = 5 on the y –5 sin t s 5 cos t s-axis. The amplitude of the motion is 5. The period of the motion is 2p. 2. The velocityy =-5 sin t attains its greatest magnitude, 5, whencos t = 0, as the graphs show in Figure 3.25. Hence, the speed of the weight, ƒ y ƒ = 5 ƒ sin t ƒ , is greatest when cos t = 0,that is, whens = 0 (the rest position). The speed of the weight is zero when t 0 3 2 5 sin t = 0.Thisoccurswhens = 5 cos t =;5, at the endpoints of the interval of motion. 2 2 2 3. The acceleration value is always the exact opposite of the position value. When the weight is above the rest position, gravity is pulling it back down; when the weight is below the rest position, the spring is pulling it back up. 4. The acceleration, a =-5 cos t, is zero only at the rest position, where cos t = 0 and FIGURE 3.25 The graphs of the position the force of gravity and the force from the spring offset each other. When the weight is and velocity of the body in Example 3. anywhere else, the two forces are unequal and acceleration is nonzero. The acceleration is greatest in magnitude at the points farthest from the rest position, where cos t =;1.

EXAMPLE 4 Jerk The jerk of the simple harmonic motion in Example 3 is da d j = = s -5 cos td = 5 sin t. dt dt

It has its greatest magnitude when sin t =;1, not at the extremes of the displacement but at the rest position, where the acceleration changes direction and sign.

Derivatives of the Other Basic Trigonometric Functions Because sin x and cos x are differentiable functions of x, the related functions sin x cos x 1 1 tan x = , cot x = , sec x = , and csc x = cos x sin x cos x sin x 4100 AWL/Thomas_ch03p147-243 8/19/04 11:16 AM Page 187

3.4 Derivatives of Trigonometric Functions 187

are differentiable at every value of x at which they are defined. Their derivatives, calcu- lated from the Quotient Rule, are given by the following formulas. Notice the negative signs in the derivative formulas for the cofunctions.

Derivatives of the Other Trigonometric Functions d stan xd = sec2 x dx d ssec xd = sec x tan x dx d scot xd =-csc2 x dx d scsc xd =-csc x cot x dx

To show a typical calculation, we derive the derivative of the tangent function. The other derivations are left to Exercise 50.

EXAMPLE 5 Find d(tan x) dx. > Solution d d cos x sin x - sin x cos x d d sin x dx dx = = A B A B Quotient Rule tan x cos x 2 dx A B dx a b cos x cos x cos x - sin x s -sin xd = cos2 x cos2 x + sin2 x = cos2 x 1 = = sec2 x cos2 x EXAMPLE 6

Find y– if y = sec x.

Solution y = sec x y¿=sec x tan x d y–= ssec x tan xd dx d d = sec x tan x + tan x sec x Product Rule dx A B dx A B = sec xssec2 xd + tan xssec x tan xd = sec3 x + sec x tan2 x 4100 AWL/Thomas_ch03p147-243 8/19/04 11:16 AM Page 188

188 Chapter 3: Differentiation

The differentiability of the trigonometric functions throughout their domains gives another proof of their continuity at every point in their domains (Theorem 1, Section 3.1). So we can calculate limits of algebraic combinations and composites of trigonometric functions by direct substitution.

EXAMPLE 7 Finding a Trigonometric Limit

22 + sec x 22 + sec 0 22 + 1 23 lim = = = =-23 x:0 cossp - tan xd cossp - tan 0d cossp - 0d -1 4100 AWL/Thomas_ch03p147-243 8/19/04 11:16 AM Page 188

188 Chapter 3: Differentiation

EXERCISES 3.4

Derivatives 26. Find y s4d = d 4 y dx4 if > In Exercises 1–12, find dy dx . =- = > a. y 2 sin x. b. y 9 cos x. =- + = 3 + 1. y 10x 3 cos x 2. y x 5 sin x Tangent Lines 1 In Exercises 27–30, graph the curves over the given intervals, together 3. y = csc x - 41x + 7 4. y = x2 cot x - x2 with their tangents at the given values of x. Label each curve and tangent with its equation. 5. y = ssec x + tan xdssec x - tan xd 27. y = sin x, -3p 2 … x … 2p 6. y = ssin x + cos xd sec x > x =-p, 0, 3p 2 cot x cos x 7. y = 8. y = > 1 + cot x 1 + sin x 28. y = tan x, -p 2 6 x 6 p 2 > > x =-p 3, 0, p 3 = 4 + 1 = cos x + x 9. y cos x 10. y x cos x > > tan x 29. y = sec x, -p 2 6 x 6 p 2 2 > > 11. y = x sin x + 2x cos x - 2 sin x x =-p 3, p 4 = 2 - - > > 12. y x cos x 2x sin x 2 cos x 30. y = 1 + cos x, -3p 2 … x … 2p > In Exercises 13–16, find ds dt . x =-p 3, 3p 2 > > > 2 13. s = tan t - t 14. s = t - sec t + 1 T Do the graphs of the functions in Exercises 31–34 have any horizontal 1 + csc t sin t tangents in the interval 0 … x … 2p? If so, where? If not, why not? 15. s = 16. s = 1 - csc t 1 - cos t Visualize your findings by graphing the functions with a grapher. 31. y = x + sin x In Exercises 17–20, find dr du. > 32. y = 2x + sin x 17. r = 4 - u2 sin u 18. r = u sin u + cos u 33. y = x - cot x 19. r = sec u csc u 20. r = s1 + sec ud sin u 34. y = x + 2 cos x In Exercises 21–24, find dp dq . > 35. Find all points on the curve y = tan x, -p 2 6 x 6 p 2, where 1 => > 21. p = 5 + 22. p = s1 + csc qd cos q the tangent line is parallel to the line y 2x. Sketch the curve cot q and tangent(s) together, labeling each with its equation. sin q + cos q tan q = 6 6 p 23. p = 24. p = 36. Find all points on the curve y cot x, 0 x , where the cos q 1 + tan q tangent line is parallel to the line y =-x. Sketch the curve and 25. Find y– if tangent(s) together, labeling each with its equation. a. y = csc x. b. y = sec x. 4100 AWL/Thomas_ch03p147-243 8/19/04 11:16 AM Page 189

3.4 Derivatives of Trigonometric Functions 189

In Exercises 37 and 38, find an equation for (a) the tangent to the 48. Is there a value of b that will make curve at P and (b) the horizontal tangent to the curve at Q. x + b, x 6 0 37. 38. gsxd = e cos x, x Ú 0 y y = = Q continuous at x 0? Differentiable at x 0? Give reasons for   your answers. 2 P  , 2  2 49. Find d 999 dx999 scos xd. > 50. Derive the formula for the derivative with respect to x of a. sec x. b. csc x. c. cot x. 1   4 P , 4 4  T 51. Graph y = cos x for -p … x … 2p. On the same screen, graph sinsx + hd - sin x x y = 0 1 2 h 2 Q y 4 cot x 2csc x for h = 1, 0.5, 0.3, and 0.1. Then, in a new window, try h =-1, -0.5, and -0.3. What happens as h : 0+ ? As x 0 1 2 3 h : 0- ? What phenomenon is being illustrated here? 4 =- - … … y 1 ͙2 csc x cot x T 52. Graphy sin x for p x 2p. On the same screen, graph cossx + hd - cos x y = h Trigonometric Limits = Find the limits in Exercises 39–44. for h 1, 0.5, 0.3, and 0.1. Then, in a new window, try h =-1, -0.5, and -0.3. What happens as h : 0+ ? As 1 1 - 39. lim sin - h : 0 ? What phenomenon is being illustrated here? x:2 ax 2 b T 53. Centered difference quotients The centered difference quotient 40. lim 21 + cos sp csc xd x: - p 6 ƒsx + hd - ƒsx - hd > p 2h 41. lim sec cos x + p tan - 1 : 4 sec x x 0 c a b d is used to approximate ƒ¿sxd in numerical work because (1) its p + tan x limit as h : 0 equals ƒ¿sxd when ƒ¿sxd exists, and (2) it usually 42. lim sin x:0 atan x - 2 sec x b gives a better approximation of ƒ¿sxd for a given value of h than Fermat’s difference quotient sin t 43. lim tan 1 - + - t:0 a t b ƒsx hd ƒsxd . h pu 44. lim cos u:0 asin u b See the accompanying figure.

y Simple Harmonic Motion Slope f'(x) = The equations in Exercises 45 and 46 give the position s ƒstd of a f(x h) f(x) Slope body moving on a coordinate line (s in meters, t in seconds). Find the C h body’s velocity, speed, acceleration, and jerk at time t = p 4 sec. B > 45. s = 2 - 2 sin t 46. s = sin t + cos t A f(x h) f(x h) Slope Theory and Examples 2h y f(x) 47. Is there a value of c that will make

sin2 3x , x Z 0 x2 ƒsxd = c, x = 0 h h x continuous at x = 0? GiveL reasons for your answer. 0 x hxx h 4100 AWL/Thomas_ch03p147-243 8/19/04 11:16 AM Page 190

190 Chapter 3: Differentiation

a. To see how rapidly the centered difference quotient for T 56. Slopes on the graph of the cotangent function Graph ƒsxd = sin x converges to ƒ¿sxd = cos x, graph y = cos x y = cot x and its derivative together for 0 6 x 6 p. Does the together with graph of the cotangent function appear to have a smallest slope? A largest slope? Is the slope ever positive? Give reasons for your sinsx + hd - sinsx - hd y = answers. 2h T 57. Exploring (sin kx)/x Graph y = ssin xd x, y = ssin 2xd x, and > > over the interval [-p, 2p] for h = 1, 0.5, and 0.3. Compare y = ssin 4xd x together over the interval -2 … x … 2. Where > the results with those obtained in Exercise 51 for the same does each graph appear to cross the y-axis? Do the graphs really values of h. intersect the axis? What would you expect the graphs of y = ssin 5xd x and y = ssin s -3xdd x to do as x : 0? Why? b. To see how rapidly the centered difference quotient for > > What about the graph of y = ssin kxd x for other values of k? ƒsxd = cos x converges to ƒ¿sxd =-sin x, graph y =-sin x > together with Give reasons for your answers. T 58. Radians versus degrees: degree mode derivatives What hap- cossx + hd - cossx - hd y = pens to the derivatives of sin x and cos x if x is measured in de- 2h grees instead of radians? To find out, take the following steps. over the interval [-p, 2p] for h = 1, 0.5, and 0.3. Compare a. With your graphing calculator or computer grapher in degree the results with those obtained in Exercise 52 for the same mode, graph values of h. sin h ƒshd = 54. A caution about centered difference quotients (Continuation h of Exercise 53.) The quotient and estimate lim : ƒshd. Compare your estimate with p 180. h 0 > Is there any reason to believe the limit should be p 180? ƒsx + hd - ƒsx - hd > 2h b. With your grapher still in degree mode, estimate cos h - 1 may have a limit as h : 0 when ƒ has no derivative at x. As a case lim . h:0 h in point, take ƒsxd = ƒ x ƒ and calculate c. Now go back to the derivation of the formula for the ƒ 0 + h ƒ - ƒ 0 - h ƒ derivative of sin x in the text and carry out the steps of the lim . h:0 2h derivation using degree-mode limits. What formula do you obtain for the derivative? As you will see, the limit exists even though ƒsxd = ƒ x ƒ has no derivative at x = 0. Moral: Before using a centered difference quo- d. Work through the derivation of the formula for the derivative tient, be sure the derivative exists. of cos x using degree-mode limits. What formula do you obtain for the derivative? T 55. Slopes on the graph of the tangent function Graph y = tan x and its derivative together on s -p 2, p 2d. Does the graph of the e. The disadvantages of the degree-mode formulas become > > tangent function appear to have a smallest slope? a largest slope? apparent as you start taking derivatives of higher order. Try it. Is the slope ever negative? Give reasons for your answers. What are the second and third degree-mode derivatives of sin x and cos x? 4100 AWL/Thomas_ch04p244-324 8/20/04 9:01 AM Page 244

Chapter APPLICATIONS OF 4 DERIVATIVES

OVERVIEW This chapter studies some of the important applications of derivatives. We learn how derivatives are used to find extreme values of functions, to determine and ana- lyze the shapes of graphs, to calculate limits of fractions whose numerators and denomina- tors both approach zero or infinity, and to find numerically where a function equals zero. We also consider the process of recovering a function from its derivative. The key to many of these accomplishments is the Mean Value Theorem, a theorem whose corollaries provide the gateway to integral calculus in Chapter 5.

4.1 Extreme Values of Functions

This section shows how to locate and identify extreme values of a continuous function from its derivative. Once we can do this, we can solve a variety of optimization problems in which we find the optimal (best) way to do something in a given situation.

DEFINITIONS Absolute Maximum, Absolute Minimum Let ƒ be a function with domain D. Then ƒ has an absolute maximum value on D at a point c if ƒsxd … ƒscd for all x in D y and an absolute minimum value on D at c if 1 y sin x ƒsxd Ú ƒscd for all x in D. y cos x

␲ ␲ x – 0 Absolute maximum and minimum values are called absolute extrema (plural of the Latin 2 2 extremum). Absolute extrema are also called global extrema, to distinguish them from local extrema defined below. –1 For example, on the closed interval [-p 2, p 2] the function ƒsxd = cos x takes on > > an absolute maximum value of 1 (once) and an absolute minimum value of 0 (twice). On FIGURE 4.1 Absolute extrema for the same interval, the function gsxd = sin x takes on a maximum value of 1 and a mini- the sine and cosine functions on mum value of -1 (Figure 4.1). [-p 2, p 2]. These values can depend Functions with the same defining rule can have different extrema, depending on the > > on the domain of a function. domain.

244 4100 AWL/Thomas_ch04p244-324 8/20/04 9:01 AM Page 245

4.1 Extreme Values of Functions 245

EXAMPLE 1 Exploring Absolute Extrema The absolute extrema of the following functions on their domains can be seen in Figure 4.2.

y y

y x2 y x2 D (–, ) D [0, 2]

x x 2 2

(a) abs min only (b) abs max and min

y y

y x2 y x2 D (0, 2] D (0, 2)

x x 2 2

FIGURE 4.2 Graphs for Example 1.

Function rule Domain D Absolute extrema on D

(a) y = x2 s - q, q d No absolute maximum. Absolute minimum of 0 at x = 0. (b) y = x2 [0, 2] Absolute maximum of 4 at x = 2. Absolute minimum of 0 at x = 0. (c) y = x2 (0, 2] Absolute maximum of 4 at x = 2. No absolute minimum.

(d) y = x2 (0, 2) No absolute extrema.

HISTORICAL BIOGRAPHY Daniel Bernoulli The following theorem asserts that a function which is continuous at every point of a (1700–1789) closed interval [a, b] has an absolute maximum and an absolute minimum value on the interval. We always look for these values when we graph a function. 4100 AWL/Thomas_ch04p244-324 8/20/04 9:01 AM Page 246

246 Chapter 4: Applications of Derivatives

THEOREM 1 The Extreme Value Theorem If ƒ is continuous on a closed interval [a, b], then ƒ attains both an absolute maximum value M and an absolute minimum value m in [a, b]. That is, there are numbers x1 and x2 in [a, b] with ƒsx1d = m, ƒsx2d = M, and m … ƒsxd … M for every other x in [a, b] (Figure 4.3).

(x2, M)

y f(x) y f(x) M M x m 1 x x ax2 b a b m Maximum and minimum at endpoints (x1, m) Maximum and minimum at interior points

y f(x)

y f(x) M M

m m x x a x2 b a x1 b Maximum at interior point, Minimum at interior point, minimum at endpoint maximum at endpoint

FIGURE 4.3 Some possibilities for a continuous function’s maximum and y No largest value minimum on a closed interval [a, b]. 1

y x The proof of The Extreme Value Theorem requires a detailed knowledge of the real 0 x 1 number system (see Appendix 4) and we will not give it here. Figure 4.3 illustrates possi- x ble locations for the absolute extrema of a continuous function on a closed interval [a, b]. 0 1 = Smallest value As we observed for the function y cos x, it is possible that an absolute minimum (or absolute maximum) may occur at two or more different points of the interval. FIGURE 4.4 Even a single point of The requirements in Theorem 1 that the interval be closed and finite, and that the discontinuity can keep a function from function be continuous, are key ingredients. Without them, the conclusion of the theorem having either a maximum or minimum need not hold. Example 1 shows that an absolute extreme value may not exist if the inter- value on a closed interval. The function val fails to be both closed and finite. Figure 4.4 shows that the continuity requirement can- x,0… x 6 1 not be omitted. y = e 0, x = 1 Local (Relative) Extreme Values is continuous at every point of [0, 1] except x = 1, yet its graph over [0, 1] Figure 4.5 shows a graph with five points where a function has extreme values on its domain does not have a highest point. [a, b]. The function’s absolute minimum occurs at a even though at e the function’s value is 4100 AWL/Thomas_ch04p244-324 8/20/04 9:01 AM Page 247

4.1 Extreme Values of Functions 247

Absolute maximum No greater value of f anywhere. Local maximum Also a local maximum. No greater value of f nearby. Local minimum y f(x) No smaller value of f nearby. Absolute minimum No smaller value of Local minimum f anywhere. Also a No smaller value of local minimum. f nearby. x a c e d b

FIGURE 4.5 How to classify maxima and minima.

smaller than at any other point nearby. The curve rises to the left and falls to the right around c, making ƒ(c)a maximum locally. The function attains its absolute maximum at d.

DEFINITIONS Local Maximum, Local Minimum A function ƒ has a local maximum value at an interior point c of its domain if ƒsxd … ƒscd for all x in some open interval containing c. A function ƒ has a local minimum value at an interior point c of its domain if ƒsxd Ú ƒscd for all x in some open interval containing c.

We can extend the definitions of local extrema to the endpoints of intervals by defining ƒ to have a local maximum or local minimum value at an endpoint c if the appropriate in- equality holds for all x in some half-open interval in its domain containing c. In Figure 4.5, the function ƒ has local maxima at c and d and local minima at a, e, and b. Local extrema are also called relative extrema. An absolute maximum is also a local maximum. Being the largest value overall, it is also the largest value in its immediate neighborhood. Hence, a list of all local maxima will automatically include the absolute maximum if there is one. Similarly, a list of all local minima will include the absolute minimum if there is one.

Finding Extrema The next theorem explains why we usually need to investigate only a few values to find a function’s extrema.

THEOREM 2 The First Derivative Theorem for Local Extreme Values If ƒ has a local maximum or minimum value at an interior point c of its domain, and if ƒ¿ is defined at c, then ƒ¿scd = 0. 4100 AWL/Thomas_ch04p244-324 8/20/04 9:01 AM Page 248

248 Chapter 4: Applications of Derivatives

Local maximum value Proof To prove that ƒ¿scd is zero at a local extremum, we show first that ƒ¿scd cannot be positive and second that ƒ¿scd cannot be negative. The only number that is neither positive nor negative is zero, so that is what ƒ¿scd must be. y f(x) To begin, suppose that ƒ has a local maximum value at x = c (Figure 4.6) so that ƒsxd - ƒscd … 0 for all values of x near enough to c. Since c is an interior point of ƒ’s domain, ƒ¿scd is defined by the two-sided limit ƒsxd - ƒscd lim - . x:c x c

Secant slopes 0 Secant slopes 0 This means that the right-hand and left-hand limits both exist at x = c and equal ƒ¿scd. (never negative) (never positive) When we examine these limits separately, we find that ƒsxd - ƒscd ¿s d = … Because sx - cd 7 0 (1) x ƒ c lim+ x - c 0. x cx x:c and ƒsxd … ƒscd Similarly, FIGURE 4.6 A curve with a local maximum value. The slope at c, ƒsxd - ƒscd ƒ¿scd = lim - Ú 0. Because sx - cd 6 0 (2) simultaneously the limit of nonpositive x:c- x c and ƒsxd … ƒscd numbers and nonnegative numbers, is zero. Together, Equations (1) and (2) imply ƒ¿scd = 0. This proves the theorem for local maximum values. To prove it for local minimum values, we simply use ƒsxd Ú ƒscd, which reverses the inequalities in Equations (1) and (2). Theorem 2 says that a function’s first derivative is always zero at an interior point where the function has a local extreme value and the derivative is defined. Hence the only places where a function ƒ can possibly have an extreme value (local or global) are 1. interior points where ƒ¿=0, 2. interior points where ƒ¿ is undefined, 3. endpoints of the domain of ƒ. The following definition helps us to summarize.

DEFINITION Critical Point An interior point of the domain of a function ƒ where ƒ¿ is zero or undefined is a critical point of ƒ.

Thus the only domain points where a function can assume extreme values are critical points and endpoints. Be careful not to misinterpret Theorem 2 because its converse is false. A differentiable function may have a critical point at x = c without having a local extreme value there. For instance, the function ƒsxd = x3 has a critical point at the origin and zero value there, but is positive to the right of the origin and negative to the left. So it cannot have a local extreme value at the origin. Instead, it has a point of inflection there. This idea is defined and discussed further in Section 4.4. Most quests for extreme values call for finding the absolute extrema of a continuous function on a closed and finite interval. Theorem 1 assures us that such values exist; Theo- rem 2 tells us that they are taken on only at critical points and endpoints. Often we can 4100 AWL/Thomas_ch04p244-324 8/20/04 9:01 AM Page 249

4.1 Extreme Values of Functions 249

simply list these points and calculate the corresponding function values to find what the largest and smallest values are, and where they are located.

How to Find the Absolute Extrema of a Continuous Function ƒ on a Finite Closed Interval 1. Evaluate ƒ at all critical points and endpoints. 2. Take the largest and smallest of these values.

EXAMPLE 2 Finding Absolute Extrema Find the absolute maximum and minimum values of ƒsxd = x2 on [-2, 1].

Solution The function is differentiable over its entire domain, so the only critical point is where ƒ¿sxd = 2x = 0, namely x = 0. We need to check the function’s values at x = 0 and at the endpoints x =-2 and x = 1: y Critical point value: ƒs0d = 0 (1, 7) Endpoint values: ƒs -2d = 4 7 ƒs1d = 1 t –2 –1 1 The function has an absolute maximum value of 4 at x =-2 and an absolute minimum y 8t t4 value of 0 at x = 0. EXAMPLE 3 Absolute Extrema at Endpoints Find the absolute extrema values of gstd = 8t - t 4 on [-2, 1].

Solution The function is differentiable on its entire domain, so the only critical points (–2, –32) ¿ = –32 occur where g std 0. Solving this equation gives 8 - 4t 3 = 0 or t = 23 2 7 1, a point not in the given domain. The function’s absolute extrema therefore occur at the FIGURE 4.7 The extreme values of endpoints, gs -2d =-32 (absolute minimum), and gs1d = 7 (absolute maximum). See gstd = 8t - t 4 on [-2, 1] (Example 3). Figure 4.7.

EXAMPLE 4 Finding Absolute Extrema on a Closed Interval 2 3 Find the absolute maximum and minimum values of ƒsxd = x > on the interval [-2, 3].

Solution We evaluate the function at the critical points and endpoints and take the largest and smallest of the resulting values. The first derivative

2 -1 3 2 ƒ¿sxd = x > = 3 323 x has no zeros but is undefined at the interior point x = 0. The values of ƒ at this one critical point and at the endpoints are Critical point value: ƒs0d = 0 2 3 3 Endpoint values: ƒs -2d = s -2d > = 24 2 3 3 ƒs3d = s3d > = 29. 4100 AWL/Thomas_ch04p244-324 8/20/04 9:01 AM Page 250

250 Chapter 4: Applications of Derivatives

y We can see from this list that the function’s absolute maximum value is 23 9 L 2.08, and it y x2/3, –2 ≤ x ≤ 3 occurs at the right endpoint x = 3. The absolute minimum value is 0, and it occurs at the interior point x = 0. (Figure 4.8). Absolute maximum; Local also a local maximum While a function’s extrema can occur only at critical points and endpoints, not every 2 maximum critical point or endpoint signals the presence of an extreme value. Figure 4.9 illustrates 1 this for interior points. We complete this section with an example illustrating how the concepts we studied x –2 –1 0 1 2 3 are used to solve a real-world optimization problem. Absolute minimum; also a local minimum EXAMPLE 5 Piping Oil from a Drilling Rig to a Refinery FIGURE 4.8 The extreme values of A drilling rig 12 mi offshore is to be connected by pipe to a refinery onshore, 20 mi = 2 3 - = ƒsxd x > on [ 2, 3] occur at x 0 and straight down the coast from the rig. If underwater pipe costs $500,000 per mile and land- = x 3 (Example 4). based pipe costs $300,000 per mile, what combination of the two will give the least expensive connection? y y x3 Solution We try a few possibilities to get a feel for the problem: 1 (a) Smallest amount of underwater pipe

x Rig –1 0 1

–1 12

(a) Refinery

20 y Underwater pipe is more expensive, so we use as little as we can. We run straight to 1 shore (12 mi) and use land pipe for 20 mi to the refinery. y x1/3 Dollar cost = 12s500,000d + 20s300,000d x –1 0 1 = 12,000,000 (b) All pipe underwater (most direct route) –1

Rig (b)

FIGURE 4.9 Critical points without ͙144 + 400 extreme values. (a) y¿=3x2 is 0 at 12 x = 0, but y = x3 has no extremum there. -2 3 Refinery (b) y¿=s1 3dx > is undefined at x = 0, > = 1 3 but y x > has no extremum there. 20

We go straight to the refinery underwater. Dollar cost = 2544 s500,000d L 11,661,900 This is less expensive than plan (a). 4100 AWL/Thomas_ch04p244-324 8/20/04 9:01 AM Page 251

4.1 Extreme Values of Functions 251

Rig

x 12 mi

Refinery

20 – y y 20 mi

Now we introduce the length x of underwater pipe and the length y of land-based pipe as variables. The right angle opposite the rig is the key to expressing the relationship between x and y, for the Pythagorean theorem gives x2 = 122 + s20 - yd2 x = 2144 + s20 - yd2 . (3)

Only the positive root has meaning in this model. The dollar cost of the pipeline is c = 500,000x + 300,000y.

To express c as a function of a single variable, we can substitute for x, using Equation (3):

cs yd = 500,0002144 + s20 - yd2 + 300,000y.

Our goal now is to find the minimum value of c(y) on the interval 0 … y … 20. The first derivative of c(y) with respect to y according to the Chain Rule is

1 2s20 - yds -1d c¿s yd = 500,000 # # + 300,000 2 2144 + s20 - yd2 20 - y =-500,000 + 300,000. 2144 + s20 - yd2 Setting c¿ equal to zero gives

500,000 s20 - yd = 300,0002144 + s20 - yd2 5 20 - y = 2144 + s20 - yd2 3 A B 25 20 - y 2 = 144 + s20 - yd2 9 A B 16 20 - y 2 = 144 9 A B 3 s20 - yd =; # 12 =;9 4 y = 20 ; 9 y = 11 or y = 29. 4100 AWL/Thomas_ch04p244-324 8/20/04 9:01 AM Page 252

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Only y = 11 lies in the interval of interest. The values of c at this one critical point and at the endpoints are cs11d = 10,800,000 cs0d = 11,661,900 cs20d = 12,000,000 The least expensive connection costs $10,800,000, and we achieve it by running the line underwater to the point on shore 11 mi from the refinery. 4100 AWL/Thomas_ch04p244-324 8/20/04 9:01 AM Page 252

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EXERCISES 4.1

Finding Extrema from Graphs 9. y 10. y (1, 2) In Exercises 1–6, determine from the graph whether the function has 5 2 any absolute extreme values on [a, b]. Then explain how your answer is consistent with Theorem 1. x 1. y 2. y –3 2 –1 y h(x) y f(x) x –2 0 2

In Exercises 11-14, match the table with a graph. x x 0 a c1 c2 b 0 a c b 11. 12. x ƒ(x) x ƒ(x) 3. y 4. y a 0 a 0 y f(x) b 0 b 0 y h(x) c 5 c 5

13. 14. x ƒ(x) x ƒ(x) x x 0 a c b 0 a c b a does not exist a does not exist b 0 b does not exist 5. y 6. y c 2 c 1.7 y g(x) y g(x)

x x 0 a c b 0 a c b abc abc

In Exercises 7–10, find the extreme values and where they occur. (a) (b) 7. y 8. y 2 1 x –1 1 x abc abc –1 –2 0 2 (c) (d) 4100 AWL/Thomas_ch04p244-324 8/20/04 9:01 AM Page 253

4.1 Extreme Values of Functions 253

Absolute Extrema on Finite Closed Intervals Local Extrema and Critical Points In Exercises 15–30, find the absolute maximum and minimum values In Exercises 45–52, find the derivative at each critical point and deter- of each function on the given interval. Then graph the function. Iden- mine the local extreme values. 2 3 2 3 2 tify the points on the graph where the absolute extrema occur, and in- 45.y = x > sx + 2d 46. y = x > sx - 4d clude their coordinates. 47.y = x24 - x2 48. y = x2 23 - x 2 15. ƒsxd = x - 5, -2 … x … 3 4 - 2x, x … 1 3 - x, x 6 0 3 49.y = 50. y = e x + 1, x 7 1 e 3 + 2x - x2, x Ú 0 16. ƒsxd =-x - 4, -4 … x … 1 -x2 - 2x + 4, x … 1 17. ƒsxd = x2 - 1, -1 … x … 2 51. y = e -x2 + 6x - 4, x 7 1 18. ƒsxd = 4 - x2, -3 … x … 1 1 2 1 15 1 - x - x + , x … 1 19. Fsxd =- , 0.5 … x … 2 52. y = 4 2 4 x2 • x3 - 6x2 + 8x, x 7 1 =-1 - … …- 20. Fsxd x , 2 x 1 In Exercises 53 and 54, give reasons for your answers. 2 3 21. hsxd = 23 x, -1 … x … 8 53. Let ƒsxd = sx - 2d > . 2 3 22. hsxd =-3x > , -1 … x … 1 a. Does ƒ¿s2d exist? 23. gsxd = 24 - x2, -2 … x … 1 b. Show that the only local extreme value of ƒ occurs at x = 2. 24. gsxd =-25 - x2, - 25 … x … 0 c. Does the result in part (b) contradict the Extreme Value p 5p Theorem? 25. ƒsud = sin u, - … u … 2 3 2 6 d. Repeat parts (a) and (b) for ƒsxd = sx - ad > , replacing 2 p p by a. 26. ƒsud = tan u, - … u … 3 4 54. Let ƒsxd = ƒ x3 - 9x ƒ . p 2p ¿s d 27. gsxd = csc x, … x … a. Does ƒ 0 exist? 3 3 b. Does ƒ¿s3d exist? p p 28. gsxd = sec x, - … x … c. Does ƒ¿s -3d exist? 3 6 d. Determine all extrema of ƒ. 29. ƒstd = 2 - ƒ t ƒ , -1 … t … 3 30. ƒstd = ƒ t - 5 ƒ , 4 … t … 7 Optimization Applications In Exercises 31–34, find the function’s absolute maximum and mini- Whenever you are maximizing or minimizing a function of a single mum values and say where they are assumed. variable, we urge you to graph the function over the domain that is ap- 4 3 propriate to the problem you are solving. The graph will provide in- 31. ƒsxd = x > , -1 … x … 8 sight before you begin to calculate and will furnish a visual context for 5 3 32. ƒsxd = x > , -1 … x … 8 understanding your answer. 3 5 33. gsud = u > , -32 … u … 1 55. Constructing a pipeline Supertankers off-load oil at a docking 2 3 34. hsud = 3u > , -27 … u … 8 facility 4 mi offshore. The nearest refinery is 9 mi east of the shore point nearest the docking facility. A pipeline must be con- Finding Extreme Values structed connecting the docking facility with the refinery. The In Exercises 35–44, find the extreme values of the function and where pipeline costs $300,000 per mile if constructed underwater and they occur. $200,000 per mile if overland. 2 3 35.y = 2x - 8x + 9 36. y = x - 2x + 4 Docking Facility 37.y = x3 + x2 - 8x + 5 38. y = x3 - 3x2 + 3x - 2 1 4 mi 39.y = 2x2 - 1 40. y = 21 - x2 Shore 1 AB Refinery 41.y = 42. y = 23 + 2x - x2 23 1 - x2 9 mi x x + 1 43.y = 44. y = x2 + 1 x2 + 2x + 2 a. Locate Point B to minimize the cost of the construction. 4100 AWL/Thomas_ch04p244-324 8/20/04 9:01 AM Page 254

254 Chapter 4: Applications of Derivatives

b. The cost of underwater construction is expected to increase, b. Interpret any values found in part (a) in terms of volume of whereas the cost of overland construction is expected to stay the box. constant. At what cost does it become optimal to construct the 60. The function pipeline directly to Point A? = + 200 6 6 q 56. Upgrading a highway A highway must be constructed to con- Psxd 2x x , 0 x , nect Village A with Village B. There is a rudimentary roadway models the perimeter of a rectangle of dimensions x by 100 x. that can be upgraded 50 mi south of the line connecting the two > villages. The cost of upgrading the existing roadway is $300,000 a. Find any extreme values of P. per mile, whereas the cost of constructing a new highway is b. Give an interpretation in terms of perimeter of the rectangle $500,000 per mile. Find the combination of upgrading and new for any values found in part (a). construction that minimizes the cost of connecting the two villages. Clearly define the location of the proposed highway. 61. Area of a right triangle What is the largest possible area for a right triangle whose hypotenuse is 5 cm long? 150 mi 62. Area of an athletic field An athletic field is to be built in the shape of a rectangle x units long capped by semicircular regions of radius r A B at the two ends. The field is to be bounded by a 400-m racetrack. 50 mi 50 mi a. Express the area of the rectangular portion of the field as a function of x alone or r alone (your choice). Old road b. What values of x and r give the rectangular portion the largest possible area? 57. Locating a pumping station Two towns lie on the south side of 63. Maximum height of a vertically moving body The height of a a river. A pumping station is to be located to serve the two towns. body moving vertically is given by A pipeline will be constructed from the pumping station to each of the towns along the line connecting the town and the pumping 1 2 s =- gt + y t + s , g 7 0, 2 0 0 station. Locate the pumping station to minimize the amount of pipeline that must be constructed. with s in meters and t in seconds. Find the body’s maximum height. 64. Peak alternating current Suppose that at any given time t (in 10 mi seconds) the current i (in amperes) in an alternating current cir- cuit is i = 2 cos t + 2 sin t. What is the peak current for this cir- 2 mi P cuit (largest magnitude)? A 5 mi Theory and Examples B 65. A minimum with no derivative The function ƒsxd = ƒ x ƒ has an absolute minimum value at x = 0 even though ƒ is not differ- 58. Length of a guy wire One tower is 50 ft high and another tower entiable at x = 0. Is this consistent with Theorem 2? Give rea- is 30 ft high. The towers are 150 ft apart. A guy wire is to run sons for your answer. from Point A to the top of each tower. 66. Even functions If an even function ƒ(x) has a local maximum value at x = c, can anything be said about the value of ƒ at x =-c? Give reasons for your answer.

50' 67. Odd functions If an odd function g(x) has a local minimum 30' value at x = c, can anything be said about the value of g at A x =-c? Give reasons for your answer. 150' 68. We know how to find the extreme values of a continuous function ƒ(x)byinvestigating its values at critical points and endpoints. But a. Locate Point A so that the total length of guy wire is minimal. what if there are no critical points or endpoints? What happens then? Do such functions really exist? Give reasons for your answers. b. Show in general that regardless of the height of the towers, the length of guy wire is minimized if the angles at A are equal. 69. Cubic functions Consider the cubic function

59. The function ƒsxd = ax3 + bx2 + cx + d. Vsxd = xs10 - 2xds16 - 2xd, 0 6 x 6 5, a. Show that ƒ can have 0, 1, or 2 critical points. Give examples models the volume of a box. and graphs to support your argument. a. Find the extreme values of V. b. How many local extreme values can ƒ have? 4100 AWL/Thomas_ch04p244-324 8/20/04 9:01 AM Page 255

4.1 Extreme Values of Functions 255

T 70. Functions with no extreme values at endpoints a. Plot the function over the interval to see its general behavior there. a. Graph the function b. Find the interior points where ƒ¿=0. (In some exercises, you 1 may have to use the numerical equation solver to approximate a sin , x 7 0 solution.) You may want to plot ƒ¿ as well. ƒ(x) = x • ¿ 0, x = 0. c. Find the interior points where ƒ does not exist. d. Evaluate the function at all points found in parts (b) and (c) and Explain why ƒs0d = 0 is not a local extreme value of ƒ. at the endpoints of the interval. b. Construct a function of your own that fails to have an extreme e. Find the function’s absolute extreme values on the interval and value at a domain endpoint. identify where they occur. T Graph the functions in Exercises 71–74. Then find the extreme values 75. ƒsxd = x4 - 8x2 + 4x + 2, [-20 25, 64 25] of the function on the interval and say where they occur. > > 76. ƒsxd =-x4 + 4x3 - 4x + 1, [-3 4, 3] > 71. ƒsxd = ƒ x - 2 ƒ + ƒ x + 3 ƒ , -5 … x … 5 2 3 77. ƒsxd = x > s3 - xd, [-2, 2] 72. gsxd = ƒ x - 1 ƒ - ƒ x - 5 ƒ , -2 … x … 7 2 3 78. ƒsxd = 2 + 2x - 3x > , [-1, 10 3] 73. hsxd = ƒ x + 2 ƒ - ƒ x - 3 ƒ , - q 6 x 6 q > 79. ƒsxd = 2x + cos x, [0, 2p] 74. ksxd = ƒ x + 1 ƒ + ƒ x - 3 ƒ , - q 6 x 6 q 3 4 1 80. ƒsxd = x > - sin x + , [0, 2p] COMPUTER EXPLORATIONS 2 In Exercises 75–80, you will use a CAS to help find the absolute extrema of the given function over the specified closed interval. Perform the following steps.