AMA211 (1) Answers to Exercise 2 (5) (6) If f ? (0) exists, hence a = 2 and b = 5. We stupefy got (a) (c) 2f(a)f ? (a); 2f ? (a); (b) (d) f ? (a); 1 ? f (a). 2 For any ?x = 0, the compound angle formula implies romaine ?x ? 1 netherworld ?x cos(x + ?x) ? cos x = cos x ? transgression x ?x ?x ?x = ?2 cos x sin ?x sin ?x sin ?x 2 . ? sin x ?x 2 ?x 2 (7) We have (a) (b) (c) (d) ?e?x sin(x ln x) + e?x [1 + ln x] cos(x ln x); 4e2x ; 2x + 1)2 (e 4x ; x4 ? 1 2x + 3 . (x2 + 3x ? 1) ln(x2 + 3x ? 1) You may then have ?x ? 0. 1 d convert x 1 d d tan x = second2 x, (ii) cot x = Ã = , (i) dx dx tan2 x dx sin2 x d d (iii) sec x = tan x sec x, (iv) csc x = ? cot x csc x, dx dx 2 3 x cos x (x + x + 10) (2x cos x ? x2 sin x) ? x2 cos x (3x2 + 1) d . = (v) 3 + x + 10 dx x (x3 + x + 10)2 (2) We have f(1 + ?x) ? f(1) = ?x f(1 + ?x) ? f(1) = lim ?x?0? ?x ?x?0+ lim 2(1 + ?x) ? 3 ? (?1) = 2 and ?x (1 + ?x) ? 2 ? (?1) lim = 1. ?x?0? ?x ?x?0+ lim (8) ?V = 15?(r + ?r)2 ? 15?r2 = 15? 2r?r + (?r)2 and dV = 30?r?r. Therefore, ?V ? dV = 15? (?r)2 . (9) We have h = f (a) = one hundred twenty + 6. When a0 = 15 and ?a = 0.01, we a 120 120 120 120 = ? ? ? ?0.0053298. a0 + ?a a0 15 + 0.01 15 then f ? (1) does not exist. (3) If ?x > 0, then ?x 3 ? 0 1 f (0 + ?x) ? f (0) = = 1 . ?x ?x ?x 3 f (0 + ?x) ? f (0) f(0 + ?x) ? f(0) = +? and lim = ??.
thence lim ?x?0+ ?x?0? ?x ?x Therefore, f ? (0) does not exist. (4) We use the filament rule to entertain f ? (x) = ? ? 1 2 1 ? (x ? µ) e? 2 ?2 (x?µ) . ? 3 2? 2 have h0 = 14 and h?h0 = ?h = f (a0 +?a)?f (a0 ) = Furthermore, f ? (a0 ) = ? 120 8 = ? . Using di?erential, we ! have a2 15 0 8 à 0.01 ? ?0.0053333. 15 ?h ? dh = f ? (a0 ) à ?a = ? (10) A simple calculation gives f ? (q) = 0.09q 2 ? 1.2q + 4.5 2 Thus f (x) = 0 if and only if x ? µ = 0. 1 and h? (q) = 250 + 90q ? 3q 2 . We thus have f ? (10) = 1.5, f ? (20) = 16.5, h? (10) = 6000 and h? (20) = 15000. Hence (13)(a) ? ?x cos x dy d à = ( x) dx dx We have 1? x ln x cos x 2 . (cosh x)2 ?...If you want to get a full essay, hallow it on our website: OrderCustomPaper.com
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