Section 2.7 Derivatives of powers of functions

Transcription

1 Section 2.7 Derivatives of powers of functions (3/19/08) Overview: In this section we iscuss the Chain Rule formula for the erivatives of composite functions that are forme by taking powers of other functions. This is a special case of the general Chain Rule which we will cover in Section 4.1. Topics: The Chain Rule for powers of functions On the orer of operations The Chain Rule for powers of functions How is the rate of change of the area A = w 2 of the square in Figure 1 etermine by its with w = w(t) an the rate of change of its with? We can view the area as the prouct A = w w an use the Prouct Rule from the last section to obtain A t = t (w2 ) = (w w) = ww t t + ww t = 2ww t. (1) This shows that the rate of change of the area equals twice the with, multiplie by the rate of change of the with. If we replace w = w(t) by f = f(x) in (1), we obtain x (f2 ) = 2f f. This is the x special case for n = 2 of the following general result: w = w(t) FIGURE 1 w = w(t) Theorem 1 (The Chain Rule for powers of functions) Suppose that n is a constant an that y = f(x) is a function of x. Then x (fn ) = nf n 1 f (2a) x or with primes enoting x-erivatives, (f n ) = nf n 1 f. (2b) This formula hols at any x such that f = f (x) exists an f = f(x) is in an open interval where [f(x)] n 1 is efine. Remember (2a) an (2b) as the following statement: the erivative of the nth power of a function equals n, multiplie by the (n 1)st power of the function, multiplie by the erivative of the function. 149

2 p. 150 (3/19/08) Section 2.7, Derivatives of powers of functions Proof: Suppose that x satisfies the conitions of the theorem. For nonzero x, we let f enote the change f(x + x) f(x) in the value of f from x to x + x, so that f(x + x) = f(x) + f. Then f 0 as x 0 since f is continuous at x. We make the simplifying assumption that f 0 for small, nonzero x. Then the erivative of y = f n at x is the limit as x 0 of [f(x + x)] n [f(x)] n x = (f + f)n f n x = [ (f + f) n f n f ][ ] f. (3) x We have written f here for f(x) an we obtaine the last expression by multiplying an iviing by f. We let x ten to zero. The ifference quotient in the secon set of brackets on the right of (3) tens to the erivative of f with respect to x an the ifference quotient in the first set of brackets tens to the erivative of f n with respect to f. The ifference quotient on the left of (3) therefore tens to the erivative of f n with respect to x, an we obtain f (fn ) = f (fn ) f x. By Theorem 1 of Section 2.4, f (fn ) = nf n 1, so the last equation gives x (fn ) = nf n 1 f x which is formula (2a) at x. QED Example 1 Fin the x-erivative of y = (x 3 + 1) 5. By (2a) with f(x) = x an n, x [(x3 + 1) 5 ] (x 3 + 1) 4 x (x3 + 1) (x 3 + 1) 4 (3x 2 ) = 15x 2 (x 3 + 1) 4. Example 2 What is z (0) if z(x) = [y(x)] 4, y(0) = 2 an y (0) = 10? Formula (2b) with y in place of f an n = 4 yiels z = x (y4 ) = 4y 3 y. We set y = y(0) = 2 an y = y (0) = 10 to obtain z (0) = 4[y(0)] 3 y (0) = 4(2) 3 ( 10) = 320.

3 Section 2.7, Derivatives of powers of functions p. 151 (3/19/08) Example 3 (a) Express the rate of change V/t of the volume V = 4 3 πr3 (4) of a sphere in terms of the raius r an the rate of change r/t of the raius. (b) At a particular moment the raius of a sphere is 3 inches an is increasing at the rate of 2 inches per minute. How fast is the volume of the sphere increasing at that moment? (a) By formula (2a) with n = 3, Example 4 V t = t (4 3 πr3 ) = 4 3 π(3r2 ) r t = 4πr2 r t. (5) (b) At the moment uner consieration, r = 3 an r/t, so that V/t = 4π(3 2 )(2) = 72π cubic inches per minute. In the next example, we use formula (5) with a graph. Figure 2 shows the graph of the volume V = V (t) of a spherical balloon as a function of the time t. Fin (a) the approximate raius of the balloon at t = 6 an (b) the approximate rate of change of the raius with respect to time at t = V (cubic inches) V = V (t) 250 V (cubic inches) V = V (t) t t (secons) (secons) FIGURE 2 FIGURE 3 (a) From the graph in Figure 2 we see that V (6) 100 cubic inches, so that by formula (4) for the volume of a sphere, πr3. This gives r 3 75/π, so that r 3 75/π. = 2.88 inches. (b) To estimate the erivative of the volume, we raw an approximate tangent line to its graph at t = 6 as in Figure 3 an fin its approximate slope. The points on the line have approximate coorinates (6, 100) an (10, 220). Using these values, we fin that the rise V from the first to the secon point on the tangent line is approximately = 120. The run t is 4. Therefore, V t V t t=6 120 cubic inches 4 secons = 30 cubic inches per secon. To fin r/t, we substitute the values V/t 30 an r 2.88 in (5). This yiels 30 4π(2.88) 2 r r an then t t 30. = inches per secon. 4π(2.88) 2

4 p. 152 (3/19/08) Section 2.7, Derivatives of powers of functions On the orer of operations Differentiating a complicate expression may require a combination of the Prouct Rule, the Quotient Rule, the Chain Rule for powers, an other operations. You can etermine the orer in which to apply these operations by noting the orer of the steps use in calculating values of the function. The ifferentiation is carrie out in the reverse orer, an this often requires the Prouct, Quotient, an Chain Rules to ifferentiate proucts, quotients, an powers of functions. ( ) 5 x + 1 Example 5 In what orer are the calculations mae in fining a value of f(x) =? The expressions y = x + 1 an y = are calculate first. Then their quotient is foun. An, finally the fifth power is taken. Example 6 Fin the x-erivative of the function y = f(x) from Example 5. x Because the last step in fining a value of y = f(x) is the taking of the fifth power, we fin its erivative by first using the Chain Rule for ifferentiating the fifth power of a function: x [ (x ) ] ( ) 4 x + 1 x ( ) x + 1. Since the next-to-last step in calculating f(x) involves ivision, the Quotient Rule is applie next: [ (x ) ] ( ) 4 x + 1 x ) 4 ( x + 1 ( ) x + 1 () (x + 1) (x + 1) x x () 2 (). (6) Finally, because the first steps in evaluating f(x) are to calculate y = x+1 an y = x 3, ifferentiating these functions gives the final result: x [ (x ) ] ( x + 1 ( x + 1 ) 4 [ ] () (x + 1) () 2 ) 4 [ ] 4 () 2 = 20(x + 1)4 () 6 It is a goo iea to write own all of the steps, as in equations (6) an (7), when you carry out involve ifferentiations such as in this example so you can concentrate on the etails an review your work to see that it is correct. (7)

5 Section 2.7, Derivatives of powers of functions p. 153 (3/19/08) Example 7 (a) Describe the orer of operations that are performe to calculate a value of g(x) = x 2 (x 3 + 2x) 10? (b) Fin the erivative of the function from part (a). Do not simplify the answer. (a) The polynomial y = x 3 + 2x is evaluate an its tenth power is taken; y = x 2 is calculate; an then the prouct is performe. (b) Because the last step in fining a value of y = g(x) is taking a prouct, the Prouct Rule is use first in fining its erivative: g (x) = x [x2 (x 3 + 2x) 10 ] = x 2 x [(x3 + 2x) 10 ] + (x 3 + 2x) 10 x (x2 ). We use the Chain Rule with the term on the left to obtain g (x) = x 2 [10(x 3 + 2x) 9 ] x (x3 + 2x) + (x 3 + 2x) 10 (2x) = 10x 2 (x 3 + 2x) 9 (3x 2 + 2) + 2x(x 3 + 2x) 10. Example 8 We o not simplify the answer because we have no further use for it. (a) What is the orer of operations in evaluating y = [x 2 +u(x)] 3/2? (b) Express the erivative of the function of part (a) in terms of x,u(x), an u (x). (a) The functions y = x 2 an y = u(x) are evaluate, the results are ae, an finally the 2 3 power is taken. (b) Since the last step in fining a value of y = [x 2 + u(x)] 3/2 is taking the power, we start with formula (2) for ifferentiating powers. We then ifferentiate y = x 2 an y = u(x): y (x) = x {[x2 + u(x)] 3/2 } = 3 2 [x2 + u(x)] 1/2 x [x2 + u(x)] = 3 2 [x2 + u(x)] 1/2 [2x + u (x)]. Interactive Examples 2.7 Interactive solutions are on the web page http// ashenk/. 1. Fin the erivative y/x of y = (10 x 1/2 ) What is W (0) if W(x) = [W(x)] 4, Z(0) = 2, an Z (0) = 10? 3. (a) Give an equation of the tangent line to y = (x 2 + 4x) 2 at x = 2. (b) Generate the curve an tangent line in a suitable winow on a calculator or computer. 4. The volume of punch in a hemispherical bowl of raius 10 inches is V = 10πh πh3 cubic inches when the punch is h inches eep (0 h 10). At what rate is the volume increasing at a moment when h is 5 inches an is increasing 6 inches per minute? In the publishe text the interactive solutions of these examples will be on an accompanying CD isk which can be run by any computer browser without using an internet connection.

6 p. 154 (3/19/08) Section 2.7, Derivatives of powers of functions Exercises 2.7 A Answer provie. CONCEPTS: O Outline of solution provie. C Graphing calculator or computer require. 1. Equation (1) states that the rate of change of the area of a square equals half its perimeter, multiplie by the rate of change of its with. Why is this plausible? (Imagine that one corner of the square is fixe.) 2. Derive Theorem 1 for n = 2 irectly from the efinition (f + f) 2 f 2 x (f2 ) = lim, where x 0 x f = f(x) an f = f(x + x) f(x) on the right. (Expan the square (f + f) 2.) 3. Derive Theorem 1 for n = 1 irectly from the efinition x (f 1 ) = lim where f = f(x) an f = f(x + x) f(x) on the right. x 0 (f + f) 1 f 1, x 4. How is the formula V t = r 4πr2 for the rate of change of the volume of a sphere from Example 3 t relate to the area A = 4πr 2 of the surface of the sphere? BASICS: Fin the erivatives in Exercises 5 through O x [(3x2 + x) 5 ] 12. O G (z) for G(z) = 3 100z A y (3) for y = (x 4 3x 3 + 1) O 3x V (0) for V = (2 4t + t 5 ) 3 x 7. O x [(a + bx2 + cx 5 ) A ] with constants a,b, c x [(3x 4) 1/3 ] 16. A x [(6 x + 3) 3 ] 8. O G (3) where G(x) = [y(x)] 7/2, y(3) = 1, an y (3) = 5 History 9. O x [(2 + 3x2 ) 10 ] 10. A y/x for y = (x + 3x 2 ) t t t [(t5 3t 2 + 1) 13 ] 18. O W (10) where W(u) = [Z(u)] 1/4, Z(10) = 16, an Z (10) = B (9) where B(v) = [A(v)] 7, A(9) = 1, an A (9) = 12 In Exercises 20 through 23 (a) give equations of the tangent lines at the given values of x. C (b) Generate the curves an tangent lines in suitable winows an copy them on your paper. 20. O The tangent line to y = (x 4 + 2) 3 at x = O The tangent line to y = (x 2 + 4x) 2 at x = The tangent line to y = 4 x at x = 3.

7 Section 2.7, Derivatives of powers of functions p. 155 (3/19/08) 23. A Figure 4 shows the graph of a ifferentiable function y = U(x). Fin approximate values of (a) U, (b) U x, (c) (xu), an () x x (U3 ) at x = 2. y y = U(x) FIGURE x 24. O The ensity of ry air at a pressure of one atmosphere an a temperature of T C is ρ(t) = 1.293( T) 1 grams per liter. (1) What is the ensity an the rate of change of the ensity with respect to temperature at 20 C? Give approximate ecimal values. 25. A A leaky bucket contains V (t) = (3 t) 2 gallons of water from t = 0 (hours) until the bucket is empty. What is the rate of flow out of the bucket when it contains 1 gallon of water? 26. What is the rate of change with respect to time of the volume V = w 3 of an expaning cubic crystal at a moment when its with is 10 millimeters an its with is increasing 3 millimetes per ay? 27. The one-imensional ensity of a ro of mass 160 grams an length L centimeters is ρ = 160/L grams per centimeter. The ro expans when it is heate. What is the rate of change with respect to temperature of its ensity when it is 40 centimeters long if its length is increasing 0.01 centimeters per egree at that time? EXPLORATION: Fin the erivatives in Exercises 28 through O y x for y = (x + 1)4 2x A t (t 2t + 3) 29. A y (x) for y = x(5x + 4) 1/ y (t) for y(t) = (t3 + 1) 10 t Give exact an approximate ecimal values, as appropriate, in Exercises 32 through A The weight of an object is the force of gravity on it. If it weighs 100 pouns on the surface of the earth, then it weighs w(r) = 100( r) 2 pouns at an altitue of r miles above the earth. (a) What oes the object weigh an (b) how rapily is its weight ecreasing when it is 400 miles above the earth if it is rising 15 miles per secon? Give exact an approximate ecimal values. 33. An object moves along the parabola y = x 2 in such a way that its x-coorinate increases at the constant rate of 5 units per minute. (a) How rapily is the object s y-coorinate increasing when its x-coorinate is 1? (b) How rapily is the object s istance to the origin ecreasing when its x-coorinate is 1? 34. A The force of air resistance (rag) on a car is D = 1 30 v2 pouns when the velocity is v miles per hour. (2) The car is accelerating at a constant rate of 500 miles per hour 2. What is the rate of increase with respect to time of the rag when the car is going 50 miles per hour? (1) CRC Hanbook of Chemistry an Physics, R. Weast eitor, Boca Raton, FL: CRC Press, Inc., 1981, p. F-11. (2) Data aopte from Flui Dynamic Drag by S. Hoerner, Publishe by the author, 1958, p. 12.

8 p. 156 (3/19/08) Section 2.7, Derivatives of powers of functions 35. Figures 5 an 6 show the graphs of the with w = w(t) (yars) an height h = h(t) (yars) of a rectangular box with a square base as functions of the time t (minutes). What is the approximate rate of change of the volume of the box with respect to t at t = 10? w (yars) w = w(t) h (yars) h = h(t) t t (minutes) (minutes) FIGURE 5 FIGURE Imagine a water pipe with a faucet at the en an with the other en connecte to a water supply that applies 50 pouns per square inch of water pressure when the faucet is close. When the valve is opene an water flows through the pipe, the water pressure in the pipe rops. The water pressure in the pipe is p 0 2v 2 pouns per square inch when the water is flowing v feet per secon. (This is an example of Bernoulli s law. ) Figure 7 shows the graph of the velocity of water as a function of t (secons) for 0 t 200. (a) What is the approximate maximum water pressure in the pipe for 0 t 200? (b) What is the approximate rate of change of the water pressure with respect to t at t = 150? FIGURE v (feet per secon) v = v(t) t (secons) 37. At what rate is the raius r of a circle ecreasing when the area of the circle is 16 square inches if the area is ecreasing 3 square inches per minute? 38. The lateral surface area of a right circular cyliner of height h (inches) an base of raius r (inches) is A = πr r 2 + h 2. Give a formula for A r h in terms of r,h,, an t t t. 39. Differentiate y = (x 3 5) 2 (a) by applying the Chain Rule for powers an (b) by expaning the square an ifferentiating the resulting expression. 40. Derive the Quotient Rule from the Prouct Rule an the Chain Rule for powers by writing f/g = fg 1. (En of Section 2.7)