Computing Derivatives With Formulas (pages 12-13), Solutions

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1 Computing Derivatives With Formulas (pages 12-13), Solutions This worksheet focuses on computing derivatives using the shortcut formulas, including the power rule, product rule, and quotient rule. We will make constant use of these techniques throughout the rest of the semester. Invest the time now that these calculations become automatic! 1. For each of the functions below, find the derivative of the function with respect to the appropriate variable. Compare your lutions with those of your classmates. Who has the best lution? (Some of these can be computed very quickly if you can find the appropriate simplification.) (a) h(x) = x 319 (b) f(s) = 4s π (c) F (x) = 1 x (d) G(s) = 3 s 5 (e) f(t) = (t + 1)(t + 2) (f) s(t) = (t 2 + 6t + 5)(t 2 2t + 9) (g) H(z) = z2 z (h) g(x) = (x 2) 5 x 2 4 Solutions: (a) Just use the power rule: (x 319 ) = 319x 318 (b) The derivative leaves the multiplier of 4 alone. Then you are left with another power rule. (4s π ) = 4(s π ) = 4πs π 1 Remember, π is just a number! If you find this calculation confusing, just think about how you would find the derivative of 4s 3.14 : (4s 3.14 ) = s 2.14 Similarly, if you include more decimal digits: (4s ) = s You would get mething similar, no matter how many digits of π you used. (c) You could use the quotient rule, but it is simpler if you rewrite the function first: F (x) = 1 x = x 1/2 F (x) = (x 1/2 ) = 1 2 x 1/2 1 = 1 2 x 3/2

2 (d) Taking a cube root is the same as raising to the 1/3 power. Using the laws of exponents, we find G(s) = 3 s 5 = (s 5 ) 1/3 = s 5/3 again we can compute the derivative using the power rule: (e) You could use the product rule: G (s) = (s 5/3 ) = 5 3 s(5/3) 1 = 5 3 s2/3 f (t) = ((t + 1)(t + 2)) = (t + 1) (t + 2) + (t + 1)(t + 2) = (t + 2) + (t + 1) = 2t + 3 Alternatively, you could have expanded the product as follows: then used the power rule: f(t) = (t + 1)(t + 2) = t 2 + 3t + 2 f (t) = (t 2 + 3t + 2) = 2t + 3 Neither method seems to be signifanctly more efficient than the other. (f) Using the product rule, we obtain ( (t 2 + 6t + 5)(t 2 2t + 9) ) = (t 2 + 6t + 5) (t 2 2t + 9) + (t 2 + 6t + 5)(t 2 2t + 9) (g) Use the quotient rule: = (2t + 6)(t 2 2t + 9) + (t 2 + 6t + 5)(2t 2) = 2t 3 4t t + 6t 2 12t t t t 2t 2 12t 10 = 4t t 2 + 4t + 44 H (z) = (z2 + 1)(z 2 ) z 2 (z 2 + 1) (z 2 + 1) 2 (z 2 + 1)(2z) z 2 (2z) (z 2 + 1) 2 = 2z (z 2 + 1) 2 (h) You could use the product rule, followed by the quotient rule. However, we can save ourself me time if we simplify first: g(x) = (x 2) 5 x 2 4 = 5(x 2) (x 2)(x + 2) = 5 x + 2 It is a lot easier to apply the quotient rule to this simplified version of g(x) : g (x) = (x + 2)(5) 5(x + 2) (x + 2) 2 = 5 (x + 2) 2 48

3 2. Assume that f(x), g(x), h(x), F (x), G(x), and H(x) satisfy the following statements. - g(3) = 17, g (3) = 2 - The tangent line to y = h(x) at x = 3 is given by equation y = 5 3(x 3) - F (x) = 3g(x) + 2h(x) - G(x) = g(x)h(x) - H(x) = h(x)/g(x) - K(x) = xg(x) Find each of the following values. (a) F (3) (b) G (3) (c) H (3) (d) K (3) Solutions: The second condition gives us h (3) = 3 (recall that h (3) is the slope of the tangent line at x = 3.) Al, the curve y = h(x) and its tangent line must agree at the point of tangency. In other words, plugging x = 3 into h(x) must give the same number as plugging x = 3 into y = 5 3(x 3), and h(3) = 5. (a): F (3) = 3g (3) + 2h (3) = 3 ( 2) + 2 ( 3) = 12 (b) requires the product rule: (c) requires the quotient rule: G (3) = g (3)h(3) + g(3)h (3) = ( 2) ( 3) = 61 (d) requires the product rule: H (3) = g(3)h (3) h(3)g (3) g(3) 2 = 17 ( 3) 5 ( 2) = K (x) = (x) g(x) + xg (x) = g(x) + xg (x) K (3) = g(3) + 3g (3) = ( 2) = 11 49

4 3. (Motivating the Product Rule) An investor currently owns n = 200 shares of a stock. Each share is currently worth s = $25. The value of the investor s portfolio at any time, t, is V (t) = n(t)s(t), where n(t) is the number of shares held at time t and s(t) is the price per share at time t. (a) Suppose the value per share holds steady, but the investor is buying stock at the rate of 50 shares per year. (i.e, n (t) = 50.) Find the rate of change of V (t). (i.e., V (t)) (b) Now suppose the value per share is increasing at the rate of $3 per share per year, but the number of shares is fixed at 200 shares. Find the rate of change of V (t). (c) Now suppose the investor is buying shares at the rate of 50 shares per year and the value of the stock is increasing at $3 per share per year. Find the rate of change of V (t). Solutions: (a) The stock is initially worth $25 per share and the investor is buying 50 additional shares per year. Thus, initially the investor is buying stock at the rate of = $1250 of stock per year. Thus, V (0) = (b) The investor initially has 200 shares of stock. Each share is gaining in value at the rate of $3 per share per year. Thus, initially the investor s stock is gaining at the rate of = $600 per year. Thus, V (0) = 600. (c) In this case, V (t) = n (t)s(t) + n(t)s (t) Initially, the stock is worth $25 per share and the investor owns 200 shares, n(0) = 200 and s(0) = 25. Al, n (t) = 50 and s (t) = 3 for all t. Thus, initally we have V (0) = n (0)s(0) + n(0)s (0) = = = Thus, the rate of change of the investor s portfolio is $1850 per year. Remark: This problem has a moral: There are two totally different ways for the value of the protfolio to change. The portfolio changes value if the number of stocks change and the portfolio al changes value if the value per share changes. Both of these can occur simultaneously. The two summands in the product rule correspond to the two different ways that the portfolio can change in value. Remark: The above argument can only be used to determine the rates of change at a fixed point in time. We can find the rate of change of V (t) at any time t as follows: The investor starts with 200 shares and adds 50 shares per year, the number of shares held at any time t is n(t) = t The stock starts at a value of $25 per share and increases $3 per year the value of a single share at time t is s(t) = t 50

5 The value of the entire portfolio is therefore V (t) = ( t)(25 + 3t) V (t) = 50(25 + 3t) + 3( t) = 300t Notice that plugging in t = 0 recovers the value V (0) = 1850 that we found above. 4. (Preview of an idea) Can you find a function f(x) whose derivative is 5x 4 7x ? Can you find more than one such function? (Hint: first find a function with derivative equal to 5x 4, then find a function with derivative equal to 7x 3, then find a function with derivative equal to 15. Now piece these functions together in an appropriate way.) Solution: We ll study this type of problem in earnest in Chapter 10. However, for this particular problem, you just need to think about the power rule in reverse. In computing the derivative of a power, the power comes out front and the power is reduced by one. Can you think of a power that when the power comes out front and the power is reduced by one, then we will be left with 5x 4? How about x 5? Check it: (x 5 ) = 5x 4 What about 7x 3? We ended up with a power of 3, we needed to start with a power of 4. But the constant of 7 doesn t seem to work out. However, we know that the derivative slides right past constant multipliers, we can modify the coeffient if needed. With me trial and error, we eventually realize (7/4)x 4 works. (An unwanted four pops out front when we take the derivative, we place a four in the denominator to intercept this unwanted four. We ll look at this systematically in chapter 10) Check that this works: ( ) 7 4 x3 = 7 4 4x3 = 7x 3 Can we find a function whose derivative is the constant 15? Why not use 15x? Thus, we see that f(x) = x x4 + 15x satisfies f (x) = 5x 4 7x Finally, the above f(x) is not the only one that works. You could have al used What other f(x) s can you find that work? f(x) = x x4 + 15x