Math150SolsFinalS11

Math150SolsFinalS11 - SOLUTIONS TO THE FINAL PART 1 MATH...

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SOLUTIONS TO THE FINAL - PART 1 MATH 150 – SPRING 2011 – KUNIYUKI PART 1: 135 POINTS, PART 2: 115 POINTS, TOTAL: 250 POINTS No notes, books, or calculators allowed. 135 points total; 3 points for each of 45 problems. You do not have to algebraically simplify or box in your answers, unless you are instructed to. DERIVATIVES (69 POINTS TOTAL) D x x 7.4 () = 7.4 x 6.4 D x x sin x = 1 sin x + x cos x = sin x + x cos x by Product Rule D x x 4 x 8 + 3 ± ² ³ ´ µ = x 8 + 3 · ¸ ¹ º » D x x 4 · ¸ ¹ º ¼ x 4 · ¸ ¹ º » D x x 8 + 3 · ¸ ¹ º x 8 + 3 2 by Quotient Rule = x 8 + 3 · ¸ ¹ º » 4 x 3 · ¸ ¹ º ¼ x 4 · ¸ ¹ º » 8 x 7 · ¸ ¹ º x 8 + 3 2 or 12 x 3 ¼ 4 x 11 x 8 + 3 2 or 4 x 3 3 ¼ x 8 x 8 + 3 2 D x ln x 6 ± ² ³ ´ µ = 6ln x 5 ± ² ³ ´ µ · D x ln x ± ² ´ µ by Gen. Power Rule = x 5 ± ² ³ ´ µ · 1 x ± ² ³ ´ µ or x 5 x D x sin 9 x ± ² ³ ´ = cos 9 x ± ² ³ ´ µ D x 9 x ± ² ³ ´ = cos 9 x ± ² ³ ´ µ 9 ± ² ³ ´ = 9cos 9 x by Gen. Trig Rule D x cos x = ± sin x D x tan x = sec 2 x D x cot x = ± csc 2 x D x sec x = sec x tan x D x csc x = ± csc x cot x
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MORE! D x e 12 x + 1 () = e 12 x + 1 ± ² ³ ´ µ D x 12 x + 1 ± ² ³ ´ = e 12 x + 1 ± ² ³ ´ µ 12 ± ² ³ ´ = 12 e 12 x + 1 D x 3 x = 3 x ln3 D x 8 x 3 = 8 x 3 ln8 ± ² ³ ´ µ D x x 3 ± ² ³ ´ = 8 x 3 ± ² ³ ´ µ 3 x 2 D x ln 2 x ± 1 ² ³ ´ µ = 1 2 x ± 1 ² ³ ´ µ · ¸ D x 2 x ± 1 ² ³ ´ µ = 1 2 x ± 1 ² ³ ´ µ · ¸ 2 ² ³ ´ µ = 2 2 x ± 1 D x log 5 x = D x ln x ln5 ± ² ³ ´ µ = 1 · D x ln x = 1 · 1 x = 1 x D x sin ± 1 x = 1 ± x 2 D x cos ± 1 x = ± 1 ± x 2 D x tan ± 1 x = 1 1 + x 2 D x sec ± 1 x = 1 xx 2 ± 1 (Assume the usual range for sec ± 1 x in our class.) D x tan ± 1 3 x ² ³ ´ µ = 1 1 + 3 x 2 ² ³ ´ µ · · ¸ D x 3 x ² ³ ´ µ = 1 1 + 3 x 2 ² ³ ´ µ · · ¸ 3 ² ³ ´ µ = 3 1 + 9 x 2 D x sinh x = cosh x D x cosh x = sinh x D x sech x = ± sech x tanh x
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INDEFINITE INTEGRALS (39 POINTS TOTAL) x 6 dx ± = x 7 7 + C 1 x dx ± = ln x + C 7 x dx ± = 7 x ln7 + C sin xdx ± = ² cos x + C cos ± = sin x + C tan ± = ² ln cos x + C or ln sec x + C cot ± = ln sin x + C sec ± = ln sec x + tan x + C csc ± = ln csc x ² cot x + C or ² ln csc x + cot x + C (Larson's variation) csc 2 ± = ² cot x + C 1 36 ± x 2 dx ² = sin ± 1 x 6 ³ ´ µ · ¸ + C 1 36 + x 2 dx ± = 1 6 tan ² 1 x 6 ³ ´ µ · ¸ + C cosh ± = sinh x + C WARNING : YOU'VE BEEN DEALING WITH INDEFINITE INTEGRALS. DID YOU FORGET SOMETHING? (I’m referring to “ + C .”)
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INVERSE TRIG FUNCTIONS (6 POINTS TOTAL) If fx () = sin ± 1 x , what is the range of f in interval form? ± ² 2 , 2 ³ ´ µ · ¸ lim x ±² tan ³ 1 x = ´ 2 (Drawing a graph may help.) HYPERBOLIC FUNCTIONS (6 POINTS TOTAL) The definition of cosh x (as given in class) is: cosh x = e x + e ± x 2 .
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Math150SolsFinalS11 - SOLUTIONS TO THE FINAL PART 1 MATH...

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