Chapter 2: Differentiation

51. What is \( \mathop{\lim }\limits_{{{\Delta x} \rightarrow 0}}\frac{\sin \left( {\frac{\pi }{4} + {\Delta x}}\right) \cos \left( {\frac{\pi }{4} + {\Delta x}}\right) - \sin \left( \frac{\pi }{4}\right) \cos \left( \frac{\pi }{4}\right) }{\Delta x} \) ?

解答

答案： ( B )

This is the limit definition of the derivative of \( f(x) = \sin x \cos x \) at \( x = \frac{\pi}{4} \). \( f'(x) = \cos^2 x - \sin^2 x \), so \( f'(\frac{\pi}{4}) = 0 \) .

52. What is \( \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{\tan \left( {\frac{\pi }{4} + h}\right) - \tan \left( \frac{\pi }{4}\right) }{h} \) ?

解答

答案： ( D )

This is \( f'(\pi/4) \) for \( f(x) = \tan x \). \( f'(x) = 1/\cos^2 x \), so \( f'(\pi/4) = 1/(\sqrt{2}/2)^2 = 2 \) .

53. If \( \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{f\left( {x + h}\right) - f\left( x\right) }{h} \) exists at every value in the domain of \( f\left( x\right) \), then: I. \( f(x) \) is differentiable. II. \( f'(x) \) exists everywhere in the domain. III. \( f(x) \) is continuous. IV. \( f'(x) \) is continuous.

解答

答案： ( C )

\( f'(x) \) may or may not be continuous; differentiability implies continuity of \( f \), so I, II, and III only.

54. What is the slope of the tangent to the curve \( 3{x}^{2} + {y}^{3} = - {37} \) when \( x = 3 \) ?

解答

答案： ( C )

Implicit differentiation: \( 6x + 3y^2 y' = 0 \) so \( y' = -2x/y^2 \). When \( x = 3 \), \( 27 + y^3 = -37 \) gives \( y = -4 \). So \( y'(3) = -6/16 = -3/8 \) .

55. What is the slope of the tangent to the curve \( \cos \left( x\right) + \frac{{y}^{2}}{2} = 1 \) when \( x = 0 \) ?

解答

答案： ( D )

When \( x = 0 \), \( 1 + y^2/2 = 1 \) so \( y = 0 \). The derivative \( y'(0) \) is undefined (vertical tangent).

56. Find \( \frac{dy}{dx} \) if \( \sec y = {\left( y - x\right) }^{3} \) .

解答

答案： ( D )

Implicit differentiation: \( \sec y \tan y \cdot y' = 3(y-x)^2(y' - 1) \). Solve for \( y' \) to get the given expression.

57. Find \( \frac{dy}{dx} \) if \( y = 7 + {5}^{{x}^{2} + {2x} - 1} \) .

解答

答案： ( D )

Chain Rule: \( \frac{dy}{dx} = (\ln 5)(5^{x^2+2x-1})(2x+2) \) .

58. The \( \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{\ln \left( {{2x} + h}\right) - \ln \left( {2x}\right) }{h} \) is

解答

答案： ( A )

This is the derivative of \( f(x) = \ln(2x) \). \( f'(x) = \frac{2}{2x} = \frac{1}{x} \) .

59. BC: The slope of the line normal to the graph of \( r = 2\cos \left( \frac{\theta }{2}\right) \) at \( \theta = \pi \) is

解答

答案： ( A )

The slope of the tangent at \( \theta = \pi \) is 1; the normal has slope \( -1 \) .

60. If \( f\left( 5\right) = 6 \) and \( {f}^{\prime }\left( 5\right) = 7 \), then the equation of the tangent to the curve \( y = f\left( x\right) \) at \( x = 5 \) is

解答

答案： ( B )

Point-slope: \( y - 6 = 7(x - 5) \) so \( y = 7x - 29 \) .

61. Table for Questions 61 and 62 given. If \( h\left( x\right) = f\left( x\right) g\left( x\right) \), then \( {h}^{\prime }\left( 2\right) = \)

解答

答案： ( A )

Product Rule: \( h'(2) = f'(2)g(2) + f(2)g'(2) = 5(-1) + 0(3) = -5 \) .

62. Using the table for Questions 61 and 62: If \( h\left( x\right) = g\left( {f\left( x\right) }\right) \), then \( {h}^{\prime }\left( 3\right) = \)

解答

答案： ( C )

Chain Rule: \( h'(3) = g'(f(3)) \cdot f'(3) = g'(1)(-2) = (-4)(-2) = 8 \) .

63. If \( 2{x}^{4} - {xy} + 3{y}^{3} = {12} \), then in terms of \( x \) and \( y \), \( \frac{dy}{dx} = \)

解答

答案： ( C )

Implicit differentiation: \( 8x^3 - y - x y' + 9y^2 y' = 0 \) so \( y' = \frac{y - 8x^3}{9y^2 - x} \) .

64. If \( f\left( x\right) = 4{\sec }^{3}\left( {5x}\right) \), then \( {f}^{\prime }\left( x\right) \) is

解答

答案： ( D )

\( f'(x) = 4 \cdot 3\sec^2(5x) \cdot \sec(5x)\tan(5x) \cdot 5 = 60\sec^3(5x)\tan(5x) \) .

65. If \( y = \frac{{e}^{3{x}^{2}}}{6} \), then \( {y}^{\prime \prime }\left( 0\right) = \)

解答

答案： ( B )

\( y' = x e^{3x^2} \), \( y'' = e^{3x^2} + 6x^2 e^{3x^2} \), so \( y''(0) = 1 \) .

66. If \( y = \frac{{2x} + 7}{5 - {2x}} \), then \( \frac{dy}{dx} = \)

解答

答案： ( A )

Quotient Rule: \( \frac{(5-2x)(2) - (2x+7)(-2)}{(5-2x)^2} = \frac{24}{(5-2x)^2} \) .

67. If \( f\left( x\right) = \ln \left( \frac{1}{x}\right) \), then \( {f}^{\prime }\left( x\right) = \)

解答

答案： ( B )

\( \ln(1/x) = -\ln x \), so \( f'(x) = -1/x \) .

68. If \( f\left( x\right) = - {\cos }^{2}\left( {{x}^{2} + {2x} - 3}\right) \), then \( {f}^{\prime }\left( x\right) = \)

解答

答案： ( C )

Chain Rule: \( f'(x) = -2\cos(\cdot)(-\sin(\cdot))(2x+2) = (4x+4)\cos(\cdot)\sin(\cdot) \) .

69. If \( y = \frac{{e}^{{x}^{2}}}{x} \), then \( {y}^{\prime } = \)

解答

答案： ( D )

Quotient and Chain: \( y' = \frac{x(2x)e^{x^2} - e^{x^2}}{x^2} = \frac{e^{x^2}(2x^2-1)}{x^2} \) .

70. What is the slope of the line tangent to \( f\left( x\right) = \ln \left( {\arcsin \left( x\right) }\right) \) at \( x = \frac{\sqrt{2}}{2} \) ?

解答

答案： ( C )

\( f'(x) = \frac{1/\sqrt{1-x^2}}{\arcsin x} \). At \( x = \sqrt{2}/2 \), \( f'(x) = \frac{4\sqrt{2}}{\pi } \) .

71. What is \( \mathop{\lim }\limits_{{{\Delta x} \rightarrow 0}}\frac{\sin \left( {\frac{\pi }{4} + {\Delta x}}\right) \cos \left( {\frac{\pi }{4} + {\Delta x}}\right) \ln \left( {\frac{\pi }{4} + {\Delta x}}\right) - \sin \left( \frac{\pi }{4}\right) \cos \left( \frac{\pi }{4}\right) \ln \left( \frac{\pi }{4}\right) }{\Delta x} \) ?

解答

答案： ( C )

This is the derivative of \( f(x) = \sin x \cos x \ln x \) at \( x = \pi/4 \). \( f'(\pi/4) = 2/\pi \) .

72. When \( {f}^{\prime \prime }\left( x\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{f\left( {x + h}\right) - f\left( x\right) }{h} \) is the definition of \( f'(x) \), then \( {f}^{\prime }\left( x\right) \) gives: I. the slope of the secant line. II. the slope of the tangent line. III. the average rate of change. IV. the instantaneous rate of change.

解答

答案： ( D )

\( f'(x) \) is the slope of the tangent line and the instantaneous rate of change (II and IV).

73. If \( f\left( x\right) = \cos \left( {3x}\right) \cdot {\sin }^{2}\left( {{2x} - \pi }\right) \), then \( {f}^{\prime }\left( \frac{\pi }{3}\right) = \)

解答

答案： ( A )

Product and Chain Rule; evaluating at \( \pi/3 \) gives \( 4(-1)(-\sqrt{3}/2)(1/2) = \sqrt{3} \) .

74. What is the slope of the tangent to the curve \( {y}^{2}\left( {{x}^{2} + {y}^{2}}\right) = 3{x}^{2} \) at \( \left( {2,\sqrt{2}}\right) \) ?

解答

答案： ( D )

Expand and use implicit differentiation; substitute \( x=2, y=\sqrt{2} \) to get \( y' = \frac{\sqrt{2}}{8} \) .

75. What is the slope of the tangent to the curve \( \sin \left( {\pi x}\right) + 9\cos \left( {\pi y}\right) = {x}^{2}y \) at \( \left( {3, - 1}\right) \) ?

解答

答案： ( C )

Implicit differentiation; at \( (3,-1) \): \( y' = \frac{\pi\cos(3\pi) - 2(3)(-1)}{9\pi\sin(-\pi)+9} = \frac{6-\pi}{9} \) .

76. Find \( \frac{dy}{dx} \) if \( {x}^{2}{y}^{2} - {3x} = 5 \) .

解答

答案： ( B )

Implicit: \( 2xy^2 + 2x^2 y y' - 3 = 0 \) so \( y' = \frac{3 - 2xy^2}{2x^2 y} \) .

77. If \( f\left( x\right) = \frac{\sqrt{x}}{2} \) and \( g\left( x\right) = \cos x \), find \( {\left[ f\left( g\left( x\right) \right) \right] }^{\prime } \) .

解答

答案： ( B )

Chain Rule: \( f'(g(x))g'(x) = \frac{1}{4\sqrt{\cos x}}(-\sin x) = -\frac{\sin x}{4\sqrt{\cos x}} \) .

78. Given the table below, find the slope of the tangent line of \( \frac{f\left( x\right) }{g\left( x\right) } \) at the point.

解答

答案： ( C )

Quotient Rule: \( \frac{f'(2)g(2) - f(2)g'(2)}{g(2)^2} = \frac{5(3)-10(7)}{9} = -\frac{55}{9} \) .

79. Given the table below, find \( {h}^{\prime \prime }\left( 3\right) \) where \( h\left( x\right) = f\left( x\right) g\left( x\right) \) .

解答

答案： ( C )

\( h'' = f''g + 2f'g' + fg'' \); at 3: \( 1(2) + 2(-5)(1) + (-1)(4) = -12 \) .

80. For \( y = - \frac{1}{4}{\log }_{2}\left( {5{x}^{2} - 9}\right) \), find \( \frac{dy}{dx} \) .

解答

答案： ( A )

\( \frac{d}{dx}[\log_2(5x^2-9)] = \frac{10x}{(\ln 2)(5x^2-9)} \); multiply by \( -1/4 \) to get \( -\frac{5x}{2\ln 2(5x^2-9)} \) .

81. For \( f\left( x\right) = \sqrt{x}{\log }_{4}\left( \frac{1}{x}\right) \), find \( {f}^{\prime }\left( x\right) \) .

解答

答案： ( C )

Product Rule and derivative of \( \log_4(1/x) \); simplifies to \( \frac{(\ln 4)\log_4(1/x) - 2}{2(\ln 4)\sqrt{x}} \) .

82. Find the slope of the tangent line to \( f\left( x\right) = \frac{{e}^{\sqrt{x}}}{x} \) at \( x = 4 \) .

解答

答案： ( D )

\( f'(x) = \frac{e^{\sqrt{x}}(\sqrt{x}/2 - 1)}{x^2} \); at \( x = 4 \), \( \sqrt{4}/2 - 1 = 0 \) so \( f'(4) = 0 \) .

83. The function \( f\left( x\right) \) is differentiable on \( \left( {1,5}\right) \) except at a nonremovable discontinuity at \( x = 3 \), with \( {f}^{\prime }\left( 1\right) = - 2 \) and \( {f}^{\prime }\left( 5\right) = 2 \) . Which is true?

解答

答案： ( D )

Both theorems require continuity (and differentiability on a closed interval); \( f \) is not continuous on \( [1,5] \) due to the discontinuity at 3.

84. For \( f\left( x\right) = {e}^{4}{\cos }^{2}\left( {5x}\right) \), find \( {f}^{\prime }\left( x\right) \) .

解答

答案： ( B )

\( e^4 \) is constant; \( (\cos^2(5x))' = 2\cos(5x)(-\sin(5x))(5) = -5\sin(10x) \); so \( f'(x) = -5e^4\sin(10x) \) .

85. Suppose \( f \) is differentiable, \( f\left( 0\right) = 2 \) and \( {f}^{\prime }\left( 0\right) = 9 \) . Find \( {h}^{\prime }\left( 0\right) \) where \( h\left( x\right) = {e}^{x}f\left( x\right) \) .

解答

答案： ( D )

Product Rule: \( h'(0) = e^0 f(0) + e^0 f'(0) = 2 + 9 = 11 \) .

86. At \( x = 3 \), \( f \) has value 2 and a horizontal tangent. If \( h\left( x\right) = {\left( f\left( x\right) \right) }^{3} \), find \( {h}^{\prime }\left( 3\right) \) .

解答

答案： ( A )

\( h'(x) = 3(f(x))^2 f'(x) \); at 3: \( h'(3) = 3(4)(0) = 0 \) .

87. For \( y = \sin \left( {x}^{3}\right) \), find \( \frac{{d}^{2}y}{d{x}^{2}} \) .

解答

答案： ( B )

\( y' = 3x^2\cos(x^3) \); \( y'' = 6x\cos(x^3) - 9x^4\sin(x^3) = 3x(2\cos(x^3) - 3x^3\sin(x^3)) \) .

88. Given \( f\left( x\right) = 3{x}^{3} + 5 \), find the slope of the tangent line to \( {f}^{-1}\left( x\right) \) at \( x = 2 \) .

解答

答案： ( D )

\( (f^{-1})'(2) = 1/f'(f^{-1}(2)) \); \( f^{-1}(2) = -1 \) since \( 3(-1)^3+5=2 \); \( f'(-1)=9 \); so slope \( = 1/9 \) .

89. If \( f\left( 3\right) = 8 \), find an expression for \( {\left( {f}^{-1}\right) }^{\prime }\left( 8\right) \) .

解答

答案： ( A )

\( (f^{-1})'(8) = 1/f'(f^{-1}(8)) = 1/f'(3) \) .

90. Evaluate \( \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{{e}^{2x} - 1}{\sin x} \) .

解答

答案： ( C )

Indeterminate 0/0; by L'Hôpital's rule: \( \lim \frac{2e^{2x}}{\cos x} = 2 \) .

91. BC: For constant \( n > 1 \), find a general formula for \( \frac{d}{dx}{\left( \frac{1}{3}\sqrt{x}\right) }^{n} \) .

解答

答案： ( D )

\( \frac{d}{dx}\left( \frac{1}{3}\sqrt{x}\right)^n = \left(\frac{1}{3}\right)^n \frac{d}{dx} x^{n/2} = \left(\frac{1}{3}\right)^n \frac{n}{2} x^{n/2 - 1} \) .

92. Suppose \( f(0) = g(0) = 0 \) for differentiable \( f \) and \( g \). The limit of \( \frac{f}{g} \) as \( x \to 0 \) is equivalent to

解答

答案： ( A )

L'Hôpital's rule: when \( f/g \) gives 0/0, \( \lim \frac{f}{g} = \lim \frac{f'}{g'} \) .

93. The first derivative of a function is linear. Which must be true of the second derivative?

解答

答案： ( D )

If \( y = mx + b \) then \( y' = m \) (constant); \( y'' = 0 \) is a constant. So second derivative is a constant, possibly zero.

94. At which values of \( x \) does \( \frac{{\sec }^{2}\left( x\right) }{4} \) have a horizontal tangent line?

解答

答案： ( B )

\( f'(x) = \frac{1}{2}\sec^2 x \tan x \); zero when \( \tan x = 0 \), i.e. \( x = n\pi \) .

95. For \( f \), the tangent line at \( x = 2 \) is \( 5x + 112 \); for \( g \), the tangent at \( x = 2 \) is \( 5x + 83 \). Which must be true?

解答

答案： ( C )

Slope of tangent = derivative; both lines have slope 5, so \( f'(2) = g'(2) = 5 \) .

96A. 96. Let \( f(x) = \frac{\sin^2 x}{x} \). (A) Find the derivative of \( f(x) \) .

解答

答案： ( )

\( f'(x) = \frac{2x\sin x\cos x - \sin^2 x}{x^2} = \frac{\sin x(2x\cos x - \sin x)}{x^2} \) .

96B. 96. (B) Write an equation for the tangent line to \( f(x) = \frac{\sin^2 x}{x} \) at \( x = \frac{\pi}{2} \) .

解答

答案： ( )

At \( x = \pi/2 \), \( f(\pi/2) = 2/\pi \) and \( f'(\pi/2) = -4/\pi^2 \). Point-slope: \( y - \frac{2}{\pi} = -\frac{4}{\pi^2}\left(x - \frac{\pi}{2}\right) \); or \( y = -\frac{4}{\pi^2}x + \frac{4}{\pi} \) .

97A. 97. Let \( f, g, h \) be differentiable. (A) Find a general formula for \( \left( \frac{fg}{h}\right)' \) .

解答

答案： ( )

\( \left(\frac{fg}{h}\right)' = \frac{(fg)'h - (fg)h'}{h^2} = \frac{(f'g + fg')h - fgh'}{h^2} \) .

97B. 97. (B) Use your formula from (a) to find \( \left( \frac{e^x\sin x}{x^2}\right)' \) .

解答

答案： ( )

\( \frac{(e^x\sin x + e^x\cos x)x^2 - e^x\sin x(2x)}{x^4} = \frac{e^x((x-2)\sin x + x\cos x)}{x^3} \) .

98. 98. Suppose \( f \) is differentiable. Find the derivative of \( h(x) = \frac{(f(x))^4}{\sqrt[3]{x}} \) .

解答

答案： ( )

Quotient Rule: \( h'(x) = \frac{4(f(x))^3 f'(x) x^{1/3} - (f(x))^4 \cdot \frac{1}{3}x^{-2/3}}{x^{2/3}} = (f(x))^3 \left( \frac{4f'(x)}{x^{1/3}} - \frac{f(x)}{3x^{4/3}}\right) = (f(x))^3 \frac{12x f'(x) - f(x)}{3x^{4/3}} \) .

99. 99. Find the equation of the tangent line to \( f(x) = \frac{x^2}{\cos x} \) at \( x = \frac{\pi}{4} \) .

解答

答案： ( )

\( f'(x) = \frac{2x\cos x + x^2\sin x}{\cos^2 x} \). \( f(\pi/4) = \frac{\pi^2\sqrt{2}}{16} \), \( f'(\pi/4) = \frac{\sqrt{2}\pi(8+\pi)}{16} \). Tangent: \( y - \frac{\sqrt{2}\pi^2}{16} = \frac{\sqrt{2}\pi(8+\pi)}{16}\left(x - \frac{\pi}{4}\right) \) or \( y = \frac{\sqrt{2}\pi(8+\pi)}{16}x + \frac{\sqrt{2}\pi^2(\pi-4)}{64} \) .

100. 100. Are there any points where both \( f(x) = \frac{3x^3+5}{\ln x} \) and \( g(x) = 4x - 9 \) have horizontal tangent lines? Justify your answer.

解答

答案： ( )

No. \( g(x) = 4x - 9 \) is linear with slope 4, so it has no horizontal tangent. Thus there is no point where both have horizontal tangents.