solHMK13 - Section 6.1 #14) If h(x) = y , then (h1 ) (y ) =...

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Section 6.1 #14) If h ( x ) = y , then ( h - 1 ) 0 ( y ) = 1 /h 0 ( x ). Hence, ( h - 1 ) 0 ( - 2) = 1 /h 0 ( - 1) = 5; ( h - 1 ) 0 (1) = 1 /h 0 (0) = 2; ( h - 1 ) 0 (4) = 1 /h 0 (1) = 5. #15) For y ( - 1 , 1), let x (0 , π ) be such that cos x = y . Then D Arccos y = 1 (cos x ) 0 = - 1 sin x = - 1 1 - cos 2 x = - 1 p 1 - y 2 . As this function blows up near the endpoints of ( - 1 , 1), it follows that Arccos y is not differentiable at y = - 1 or y = 1. #16) For y R , let x ( - π/ 2 , π/ 2) be such that tan x = y . Then D Arctan y = 1 (tan x ) 0 = 1 sec 2 x = 1 1 + tan 2 x = 1 1 + y 2 . #17) Let ± > 0. Choose δ > 0 such that if 0 < | x - c | < δ , then ± ± ± f ( x ) - f ( c ) x - c - f 0 ( c ) ± ± ± < ± . Thus, for 0 < | x - c | < δ , we have | f ( x ) - f ( c ) - f 0 ( c )( x - c ) | < ± | x - c | . Clearly, if x = c , then both sides are 0, so for | x - c | < δ , we have | f ( x ) - f ( c ) - f 0 ( c )( x - c ) | ≤ ± | x - c | . Let c - δ < u c v < c + δ . Then | f ( u ) - f ( c ) - f 0 ( c )( u - c ) | ≤ ± | u - c | = ± ( c - u ) | f
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solHMK13 - Section 6.1 #14) If h(x) = y , then (h1 ) (y ) =...

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