hw2.ms (2713B)
1 .TL 2 Calculus assignment 2 3 .AU 4 Lucas Standen (lus53@aber.ac.uk) 5 6 .EQ 7 delim @@ 8 .EN 9 .EQ 10 delim @# 11 .EN 12 13 .2C 14 15 .LP 16 3)a) @f sub 1 (x) = x sup 2# 17 18 @f sub 2 (x) = sqrt x# 19 20 @f sub 1 (1) = 1# 21 22 @f sub 2 (1) = 1# 23 24 @f sub 1 (1) = f sub 2 (1)# 25 26 Therefore f(x) is continuous 27 28 @f' sub 1 (x) = 2x# 29 30 @f' sub 2 (x) = 1 over {2 sqrt x}# 31 32 @lim sub { -> 1} f' sub 1 (x) = 1# 33 34 @lim sub {x -> 1} f' sub 2 (x) = 1 over 2# 35 36 @lim sub {x -> 1} f' sub 2 (x) != lim sub { -> 1} f' sub 1 (x)# 37 38 Therefore f(x) is not differentiable 39 40 3)b) 41 .I 42 See end of doc. 43 44 5)c) @lim sub {h -> 0} {sqrt {3 (x + h) + 1} - sqrt {3 x + 1}} over h# 45 46 @lim sub {h -> 0} {sqrt {3 (x + h) + 1} - sqrt {3 x + 1}} over h {sqrt {3 (x + h) + 1} + sqrt {3 x + 1}} over {sqrt {3 (x + h) + 1} + sqrt {3 x + 1}}# 47 48 @lim sub {h -> 0} {(3 (x + h) + 1) - {(3 x + 1)}} over { h sqrt {3 ( x + h ) + 1} + sqrt {3 x + 1}}# 49 50 @lim sub {h -> 0} { 3h } over { h sqrt {3 ( x + h ) + 1} + sqrt {3 x + 1}}# 51 52 @lim sub {h -> 0} { 3 } over { sqrt {3 ( x + h ) + 1} + sqrt {3 x + 1}}# 53 54 @3 over { 2 sqrt {3 x + 1}} # 55 56 6)b) @u = sqrt x# 57 58 @u' = 1 over { 2 sqrt x }# 59 60 @v = sin x# 61 62 @v' = cos x# 63 64 @ dy over dx = 1 over {2 sqrt x} sin x + sqrt x cos x# 65 66 6)c) @u = 2x# 67 68 @u' = 2# 69 70 @v = 4 + x sup 2# 71 72 @v' = 2x# 73 74 @ dy over dx = {{2 (4 + {x sup 2})} - {4 x sup 2}} over {(4 + {x sup 2})} sup 2# 75 76 @ dy over dx = - {2x sup 2 + 8} over {(4 + {x sup 2})} sup 2# 77 78 7)d) 79 80 let @f(x) = {x sup 2 + 1} over {x sup 2 - 1}# 81 82 let @y = f(x) sup 3# 83 84 @f' (x)# 85 86 @u = x sup 2 + 1# 87 88 @u' = 2x# 89 90 @v = x sup 2 - 1# 91 92 .I 93 next page, first column 94 95 .LP 96 @v' = 2x# 97 98 @f'(x) = {{2x (x sup 2 - 1)} - {2x (x sup 2 + 1)}} over {(x sup 2 - 1) sup 2}# 99 100 @f'(x) = {{(x sup 3 - 2x)} - {(x sup 3 + 2x)}} over {(x sup 2 - 1) sup 2}# 101 102 @f'(x) = {-4x} over {(x sup 2 - 1) sup 2}# 103 104 @dy over dx = 3 f(x) sup 2 f'(x)# 105 106 @dy over dx = -12 ({{{x sup 2} + 1} over {{x sup 2} - 1}}) sup 2 x over {{ (x sup 2 - 1) } sup 2}# 107 108 8) @g'(x) = cos x - 8 sin 4x# 109 110 @g''(x) = -sin x - 32 cos 4x# 111 112 @g'({pi over 4}) = {sqrt 2} over 2# 113 114 @g''({pi over 4}) = {64 - sqrt 2} over 2# 115 116 9)c) 117 Differentiate the first half 118 @d over dy cos(x + y)# 119 120 @y = cos t# 121 122 @y' = -sin t# 123 124 @t = x + y# 125 126 @t' = 1 + dy over dx# 127 128 @dy over dx = t' sin t# 129 130 @dy over dx = -{({1 + dy over dx})}(sin(x + y))# 131 132 133 Differentiate the second half 134 @d over dy sin(x + y)# 135 136 @y = sin t# 137 138 @y' = cos t# 139 140 @t = x + y# 141 142 @t' = 1 + dy over dx# 143 144 @dy over dx = t' sin t# 145 146 @dy over dx = {({1 + dy over dx})}(cos(x + y))# 147 148 Put the whole thing together 149 150 @1 over 3 = (1 + {dy over dx}) (-sin (x + y) + cos (x + y))# 151 152 @1 over {3 (1 + {dy over dx})} = (-sin (x + y) + cos (x + y))# 153 154 @1 over {3 (-sin (x + y) + cos (x + y)) } = (1 + {dy over dx}) # 155 156 @1 over {3 (cos (x + y) - sin(x + y)) } - 1 = {dy over dx} # 157 158 .PSPIC graph.ps