Computing Derivatives With Formulas Some More (pages 14-15), Solutions

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1 Computing Derivatives With Formulas Some More pages 14-15), Solutions This worksheet focuses on computing derivatives using the shortcut formulas, including the power rule, product rule, quotient rule, chain rule, exponential functions. 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) hx) = x + 5x 2 b) fs) = s s + 7 c) 5 F x) = x 2) x 2 4 d) Gx) = xe x2 e) ft) = e 5t2 3t+1 f) st) = t + 1)t + 2) g) Hz) = ze z h) gx) = e 8x Solutions: a) Use the chain rule: Inside = x + 5x 2 Outside = ) 1/2 Inside = x Outside = 1 ) 1/2 2 h x) = 1 2 x + 5x 2 ) 1/ x) b) Use the product rule. s s + 7) 1/2 ) = s s + 7) 1/2 + s s + 7) 1/2) c) It is best to simplify first: Now use the quotient rule: s + 7) 1/2 + s 1 2 s + 7) 1/2 = s + 7) 1/2 + s s + 7) 1/2 2 F x) = x 2) Alternatively, we could have written F x) as 5 x 2 4 = x 2) 5 x 2)x + 2) = 5 x + 2 F x) = x + 2)5 5x + 2) x + 2) 2 = 5 x + 2) 2 F x) = 5x + 2) 1,

2 then used the chain rule inside function equal to x + 2, outside function equal to ) 1 ) F x) = 5 1)x + 2) 2 = 5 x + 2) 2 d) This requires the product rule the chain rule. First lets just worry about the e x2 term: Here Inside = x 2 Outside = e ) Now, e) Use the chain rule Inside = 2x Outside = e ) e x2 ) = 2x e x2 G x) = x e x2 + xe x2 ) = e x2 2x 2 e x2 f) Multiply out first, then use the power rule: Inside = 5t 2 3t + 1 Outside = e ) Inside = 10t 3 Outside = e ) f t) = 10t 3) e 5t2 3t+1 st) = t + 1)t + 2) = t 2 + 3t + 2 s t) = 2t + 3 Alternatively, you could have used the product rule) g) Use the product rule the chain rule. Lets focus on e z first: Inside = z Outside = e ) Inside = 1 2 z 1/2 Outside = e ) e ) z 1 = 2 z 1/2 e z Now, H z) = z e z ) + z e z) = e z 1 + z 2 z 1/2 e z 53

3 = e z z1/2 e z h) This requires the chain rule: Inside = 8x Outside = e ) Inside = 8 Outside = e ) e 8x ) = 8 e 8x 2. Discussion: Suppose f2) = 3, f 2) = 4, g1) = 3, g 1) = 5, g2) = 4 g 2) = 5. Suppose F x) = f2x) gx). Find F 1) = 1. Is the answer 1, or 1 3? Solution: The quotient rule is required to compute F x). What may be less obvious is that the chain rule is al required. In what follows, I ll write f2x)) to denote the derivative of the function f2x). This is not the same as f 2x). The latter means the derivative of the function fx) evaluated at 2x. The function f2x) consists of an inside function namely, 2x) an outside function namely, f )), f2x)) = 2 f 2x) Plugging in x = 1 therefore gives 2 f 2) = 2 4 = 8. Now, F x) = gx)f2x)) f2x)g x) gx) 2 Plugging in x = 1 gives = gx) 2 f 2x) f2x)g x) gx) 2 F 1) = g1) 2 f 2) f2)g 1) g1) 2 = = 1 3. For each of the following functions, determine x that f x) = 0. a) fx) = xe x b) fx) = xe 4x c) fx) = xe 7x 54

4 d) fx) = xe rx where r is an arbitrary constant. Solutions: d) The abstract case subsumes the three previous cases: Setting this to zero: f x) = x) e rx + xe rx ) = e rx + xe rx rx) = e rx rxe rx = 1 rx)e rx 1 rx)e rx = 0 So either 1 rx = 0 or e rx = 0. But e mething can never be zero, we must have 1 rx = 0, x = 1/r. 4. Let fx) = xe x. Find a) f x) b) f x) c) f 328) x) Use parts a b, compute the third, fourth, etc. derivatives until you notice the pattern.) Solutions: Use the product rule for a): f x) = e x xe x f x) = e x xe x) = e x ) xe x ) But e x ) = e x we just saw that xe x ) = e x xe x, Taking another derivative, we find Taking another derivative, we find f x) = e x e x xe x ) = 2e x + xe x f x) = 2e x + xe x) = 2e x + e x xe x ) = 3e x xe x f 4) x) = 3e x xe x) = 3e x + e x xe x ) = 4e x + xe x It appears that each derivative involves two terms: one term is e x multiplied by the number of times the derivative has been applied, the other term is xe x. Al, one of the terms is positive the other is negative, each time we take a derivative the negative sign switches to the other term. Thus, we expect f 328) x) = 328e x + xe x If you know about mathematical induction, you could try to prove a general formula for the higher order derivatives of xe x.) 55

5 5. Let fx) = e 0.5x. a) Find the equation of the tangent line to y = fx) at x = 0. b) On the same set of axes, draw the graph of y = fx) the tangent line found above. c) Suppose you need a good approximation to f0.1) = e 0.05, but your trusty calculator is not available. Explain how you might use your answers to parts a b to find such an approximation. d) Now use a calculator to evaluate e How good is your approximation in part c)? Solutions: a) Point on the tangent line is 0, f0)) = 0, 1). The slope of the tangent line is f 0) = 0.5. The tangent line is therefore y 1 = 0.5x 0) = y = 0.5x + 1 b) c) The tangent line the curve meet exactly at x = 0. At x = 0.1 the tangent line the curve do not agree, but they are extremely close. Therefore, plugging x =.1 into y = 0.5x + 1 plugging x = 0.1 into y = e 0.5x should give very similar values. Thus, we expect f0.1) = e = 1.05 How good is this approximation? Using a caclulator, we find e 0.05 = our approximation is within ± of the correct value. In particular, our approximation is correct to three significant figures. 56

6 Remark: This rt of approximation is used quite often in finance. In most financial situations, interest is credited on a compound basis Using A = P e rt. Over short periods of time, or when the interest rate is very low, simple interest i.e., A = P 1 + rt)) is used as a simple approximation to compound interest. The above calculation or a more general form of the above calculation) justifies using simple interest as an approximation to compound interest. 57