Maths with Maple Week 5 tutorial. Differentiation 1. dx ln(x) ln(x) 2. = 2x ln(x) x2.x 1. ln(x) x

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1 Maths with Mapl W 5 tutorial Diffrntiation Solution. Th quotint rul givs. ln. ln ln ln ln ln. ln. ln Not hr that ln is not th sam as ln ln. For ampl, ln, ln, but ln. Solution. Put u a + b an v c +, u a an v c. W thn hav y u/v u v uv /v ac + a ac + bc/c + a bc c +. Solution 3. Put u a + b/ an v c + / an y u/v; w must fin y. Not that u a b/ v c / u v uv a b/ c + / a + b/c / ac + a/ bc/ b/ 3 ac + a/ bc/ + b/ 3 a bc/, y u v uv a bc v c + /. Solution 4. First, rcall that cos t sint, t + a cost a sint. Using this an th quotint rul, w gt t sint + a cost cost a sint sint a sint cost + a sint cost + a cost + a cost sint + a cost. Nt, w hav + y cost + sint + a cost + a cost, + y + a cost. Multiplying ths two rsults togthr, w gt + y t sint. Solution 5. W us th prouct rul, an th fact that. This givs Th nt thing in th squnc is

2 Th cofficints hr ar, 5, 5 4 0, , an Solution 6. Th most fficint mtho is as follows: Altrnativly, w hav y /y lny lnp + lnq lnr lns p /p + q /q r /r s /s. y pq rs pqrs rs p qrs + pq rs pqr s pqrs r s, y y p qrs + pq rs pqr s pqrs r s p qrs + pq rs pqr s pqrs pqrs p /p + q /q r /r s /s. Now ta p, q + 3, r + an s +, y + 3/ + +. W thn hav p q r s, rs pq y Solution 7. a,b Th lft han pictur shows a typical o function f. Th tangnt lins at an hav th sam slop, f f, f is vn. f gg f f Th right han pictur shows a typical vn function g. Th tangnt lins at an hav opposit slops, g g, g is an o function. c, Ta f +. This is nithr vn nor o bcaus f an f 0 which is nithr + nor. Howvr, f is th constant function, which is vn. Thus this f answrs both c an. Solution 8. Probably th simplst answr is f cosπ, f π sinπ, f n 0 for all intgrs n. W al hav f0 an f, f0 < f as rquir. Anothr approach is to ta f / 4 /4, f 3 +, again f 0 f f 0. In this cas w hav f0 0 an f /4, f0 < f.

3 3 Solution 9. Th obvious ampl is f f < 0. Othr ampls inclu + an tanh. Solution 0. W hav f a + b for all. As f 0 0, w must hav a 0 + b 0, 0 b/a. W al hav f 0 0, 0 b/a ± b 4ac/a. Ths two prssions for 0 can only b compatibl if b 4ac/a 0, b 4ac 0, c b /4a. Solution. 3/ g v v c v v c powr rul 3/ v v/c c 3/ vc v c. Solution. Put t a an u pt a an v sinω an y uv a sinω; w must fin y/. W hav t a u t t t t a u t u t a a v ω cosω y u v + u v a ω cosω a a sinω a ω cosω a sinω. It is a common mista for things li a to crp in. If a wr a variabl thn w woul hav a a a. Howvr, a is in fact a constant, an w ar using rathr than a, th quation a a a is not rlvant. Solution 3. Put u p q, y u /pq. Thn u/ p p q q p p q q an y u pq u/pq pq p y q /pq. W thrfor hav p q y y u u u p q pq p q /pq p p q q p q /pq p p q q /pq p q /pq p /q q /p. Solution 4. Put u a/ b an y f u n. Thn u. b a. b a b b, f y n u a nun na b b na b a n b n. b

4 4 Solution 5. It is convnint to introuc th notation Lz z. Th qustion thn ass us to fin Ly, LLy an LLLy. W hav Ly y a + a + 3a 3 + 4a 4 3 a + a + 3a a 4 4 LLy a + a + 3 3a a 4 3 a + 4a + 9a a 4 4 a + a + 3 a a 4 4 LLLy a + 4a + 3 9a a 4 3 a + 8a + 7a a 4 4 a + 3 a a a 4 4. Th gnral rul is clarly that if y a, thn L n y n a. Solution 6. W first calculat th succssiv rivativs: y y + y It follows that y y + y + y + y y + y + y 3 3 y y + y + y + y + y + y y + 3 y + 3 y + y. y y + y y y + y + y 3 3 y y + 3y + 3y + y. You shoul rcogniz th numbrs hr as binomial cofficints; thy ar th sam as in th formula Th pattrn sms to b that n + t + t n y y + n + t + t + t + t 3 + 3t + 3t + t 3. 0 y n + y. y + + n y n On way to prov this is by inuction; w omit th tails. Hr is anothr, mor abstract approach; you can ignor it if you ar not intrst. Consir th oprators Dz z an Lz z. W hav sn that Lz z + z + Dz, L + D, L n + D n n 0 D,

5 5 On th othr han, w hav an on. It follows that as claim. Solution 7. L n y D y L y LLy Ly y L 3 y + / In othr wors, w hav c 3/. Solution 8. y 3 3 y n n y L n y + + y. y y,. + /. + /. + + / + / +, /. + f / g h 7 tan 6 tan 7 tan 6 + tan 6 sin 7 cos cos 7 sin7 / cos 9. / sin arcsin / cosarcsin / m ln cos cos sin. sin cos cos tan.