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MATH 151 Midterm: MATH 151 TAMU Y2013 2013a Exam 2a Solutions

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turquoisegnu128
31 Jan 2019
School
Texas A&M University
Department
Mathematics
Course
MATH 151
Professor
All

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Document Summary

Exam ii version a solutions: d v = s = 2t 1 = 0 when t = 2(cid:19) f (0)(cid:12)(cid:12)(cid:12)(cid:12) eled is (cid:12)(cid:12)(cid:12)(cid:12) f(cid:18) 1 (cid:12)(cid:12)(cid:12)(cid:12) 2(cid:19)(cid:12)(cid:12)(cid:12)(cid:12) f (2) f(cid:18) 1 ft . = 2: a di erentiate implicitly: 3y2 dy dy dx dy dx. 7 (3y2 + 7) = 2x, dy dx dx 2x = 3y2 + 7: b f (4)(x) = f (x) = cos x, so the derivatives repeat every four. Since 2013 4 has a re- mainder of 1, f (2013)(x) = f (x) = sin x. f (x)g (x) g(x)f (x) U (x: e (f (x))2 so u (1) = (2)(2) ( 1)(3) 4: d v (x) = g(x)f (x)+f (x)g (x), so v (1) = g(1)f (1) + f (1)g (1) = ( 1)(3) + (2)(2) = 1, b s (x) = 4(f (x))3(f (x)), so s (1) =

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