Solução_Calculo_Stewart_6e

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F. TX.10 3 DIFFERENTIATION RULES 3.1 Derivatives of Polynomials and Exponential Functions e h − 1 1. (a) e is the number such that lim =1. h→0 h (b) x 2.7 x − 1 x −0.001 0.9928 −0.0001 0.9932 0.001 0.9937 0.0001 0.9933 x 2.8 x − 1 x −0.001 1.0291 −0.0001 1.0296 0.001 1.0301 0.0001 1.0297 From the tables (to two decimal places), 2.7 h − 1 2.8 h − 1 lim =0.99 and lim =1.03. h→0 h h→0 h Since 0.99 < 1 < 1.03, 2.7
F. 88 ¤ CHAPTER 3 DIFFERENTIATION RULES TX.10 27. We first expand using the Binomial Theorem (see Reference Page 1). H(x) =(x + x −1 ) 3 = x 3 +3x 2 x −1 +3x(x −1 ) 2 +(x −1 ) 3 = x 3 +3x +3x −1 + x −3 ⇒ H 0 (x) =3x 2 +3+3(−1x −2 )+(−3x −4 )=3x 2 +3− 3x −2 − 3x −4 29. u = 5√ t +4 √ t 5 = t 1/5 +4t 5/2 ⇒ u 0 = 1 5 t−4/5 5 +4 2 t3/2 = 1 5 t−4/5 +10t 3/2 or 1/ 5 5√ t 4 +10 √ t 3 31. z = A y + 10 Bey = Ay −10 + Be y ⇒ z 0 = −10Ay −11 + Be y = − 10A y + 11 Bey 33. y = 4√ x = x 1/4 ⇒ y 0 = 1 4 x−3/4 = 1 4 4√ x . At(1, 1), 3 y0 = 1 4 and an equation of the tangent line is y − 1= 1 4 (x − 1) or y = 1 4 x + 3 4 . 35. y = x 4 +2e x ⇒ y 0 =4x 3 +2e x . At (0, 2), y 0 =2and an equation of the tangent line is y − 2=2(x − 0) or y =2x +2. The slope of the normal line is − 1 (the negative reciprocal of 2) and an equation of the normal line is 2 y − 2=− 1 (x − 0) or y = − 1 x +2. 2 2 37. y =3x 2 − x 3 ⇒ y 0 =6x − 3x 2 . At (1, 2), y 0 =6− 3=3, so an equation of the tangent line is y − 2=3(x − 1) or y =3x − 1. 39. f(x) =e x − 5x ⇒ f 0 (x) =e x − 5. Notice that f 0 (x) =0when f has a horizontal tangent, f 0 is positive when f is increasing, and f 0 is negative when f is decreasing. 41. f(x) =3x 15 − 5x 3 +3 ⇒ f 0 (x) =45x 14 − 15x 2 . Notice that f 0 (x) =0when f has a horizontal tangent, f 0 is positive when f is increasing, and f 0 is negative when f is decreasing.
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