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**Unformatted text preview: **Math 106 Exam 1 February 20, 2001 R. Durrett
(4 points) 0. Write your name and section number on your exam book. (16 points) 1. Find the derivative of 2 m3+6$+3\/‘——-2—
cc (60 points = 10 per part) 2. Differentiate the following (a) m2 sinzc ((1) 337(1 + 2m3)4 5:32—23“ . (e) sinlacncaosm (C) 1n(1 + 3332) (f) w/l + BSin(:c4) (20 points) 3. Find the equation of the tangent line to f(:z:) = 331/3 at
:c r: 216. Use the tangent line to estimate (250)1/3. Math 106 Answers to Exam 1. 2/20/2001 R. Durrett
1. (m3 + 6:): + 3331/2 — 2m"2)' 2' 3502 + 6 +gw‘1/2 + 41:”3
2- a« Since (f9)' = f'g + f9':
(3:2 sinx)’ = 230 since + 3:2 cos a:
b. Since (f/g)’ = (f’g ~ fg’)/g2,
(6223/:172)’ : (26236322 - e2$2m)/334 G. Since f(g)’ = f'(9)9', (111(1 + 3332” - 1 _ .6
1+3$2 ‘6 d. Since (fg)’ = f’g+f9’7 (587(1 + 2x3)4)' = 73:6(1 + 2x3)4 + 337 ~ 4(1 + 2$3)36x2 e. Since (f/g)’ = (fig ’ f9/)/92, 233 —— sin2 :13)1n$ —— sinxcosatﬂ/ac) (In as)2 (sinascosa:)/1nx)' = (COS f. Since f(g)’ = f’(9)g’, H 1 + 3 sin(a:4), %(1 + 3 sin x4)"1/23cos(m4)4x3
3. If ms) = 5171/3 then f’(:c) = (1/3):c-2/3, f(216) = 6, f’(216)= 1/(3 - 36). The equation of the tangent line is
1
1—0—8 At a: = 250 this is 6—55—48 = 6.3148 compared to (250)1/3 = 6.2996. y=6+ (23—216) ...

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