HW2-510-soln - AMS 510 HW2 Solution 4.39 3 (a) 0 f x = x...

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AMS 510 HW2 Solution 4.39 (a) 0 ≤∣ f x ∣= x 3 sin 1 x ≤∣ x 3 , as lim x 0 x 3 = 0 lim x 0 f x ∣= 0 , that is lim x 0 f x = 0 = f x , so it is continuous. (b) lim h 0 f 0 h − f 0 h 0 = lim h 0 h 3 sin 1 h h 0 = lim h 0 h 2 sin 1 h , from h 2 sin 1 h ≤∣ h 2 and lim h 0 h 2 = 0 shows that lim h 0 h 2 sin 1 h = 0 or lim h 0 h 2 sin 1 h = 0 , which indicates f ' 0 = lim h 0 f 0 h − f 0 h 0 = 0 . (c) as f ' x = { 3 x 2 sin 1 x – x cos 1 x x 0 0 x = 0 , f ' x ∣= 3x 2 sin 1 x – x cos 1 x 3x 2 sin 1 x x cos 1 x 3x 2 ∣ x , lim x 0 3 x 2 ∣ x ∣= 0 gives lim x 0 f ' x ∣= 0 or lim x 0 f ' x = 0 = f ' 0 , which indicates f ' x is continuous at x = 0 . 4.56 (a) from the Chain Rule, d a u x dx = d a u x du du dx = a u x ln a du dx (b) d cscu dx = d csc u du du dx =− csc u cot u du dx (c) d tanh u dx = d tanh u du du dx = sech 2 u du dx 4.57 (a) y = tan 1 x x = tan y
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dy dx = 1 dx dy = 1 sec 2 y = 1 1 tan 2 y = 1 1 x 2 (b) y = csc 1 x x = csc y , cot y =± x 2 1 dy dx = 1 dx dy = 1 csc y cot y = 1 x ⋅± x 2 1 =∓ 1 x x 2 1 (c) y = sinh 1 x x = sinh y dy dx = 1 dx dy = 1 cosh y = 1 1 sinh 2 y = 1 1 x 2 (d) y = coth 1 x x = coth y dy dx = 1 dx dy = 1 1 sinh 2 y =− sinh 2 y = 1 1
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This note was uploaded on 01/31/2011 for the course AMS 510 taught by Professor Feinberg,e during the Fall '08 term at SUNY Stony Brook.

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HW2-510-soln - AMS 510 HW2 Solution 4.39 3 (a) 0 f x = x...

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