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1、2005/10/7Engineering Mathematics calculus1Calculus for Engineering Mathematics 微積分複習-導的定義1. 導的定義:為此曲線之斜為函之曲線，則之一階導為)( )()()( )()()(lim 0xfxfxfxfxxfxxfxfx+=如速v(t)為位移 r(t)之一次微分，加速a(t)為位移 之二次微分( )22dtd dtddtdtrvarvrr=2005/10/7Engineering Mathematics calculus2Engineering Mathematics 微積分複習-基本微分公式2.基本微分公式
2、 a.設n為正整,則b.c為一常,則c.n為正整,則或1=nn nxdxdx0=dxdcnnnxndxxd1*1=111*1=nn xndxdx2005/10/7Engineering Mathematics calculus3Engineering Mathematics 微積分複習-微分公式四則運算)()( *)()( *)()()()()(.)()()()()()()()(.)(*)()(*)(.)()()()()()(. 32xvxvxuxuxvxyxvxuxydxvxuxvxuxyxvxuxycxucxyxucxybxvxuxyxvxuxya=+=微分公式四則運算：2005/10/7
3、Engineering Mathematics calculus4Engineering Mathematics 微積分複習-微分公式Sample:3221334 32 3231 3232 32)2)(1(21 )2(1*) 1(1*)2(*)21(21)()(21)(. 21228) 12(26*) 12(32*)()() 12()(. 1+=+ +=+=+=xxxxx xxxfxfxxxfxxxxxxxxxfxfxxxf，則設，則設2005/10/7Engineering Mathematics calculus5Engineering Mathematics 微積分複習-初等函微分公式4
4、.初等函的導公式 a.b.c.d.xxee= )()(1)(ln證明下一頁xx=axxaln1)(log=aaaxxln)(=xxxxxxxxxxxxxxcot*csc)(csctan*sec)(seccsc)(cotsec)(tansin)(coscos)(sin22=2005/10/7Engineering Mathematics calculus6故得證最後可得而又移至等號另一側將此式簡化後可得可寫成依微分等號側同時對則等號側同時取證明xxdxdxexyedxdyedxdyexdxd dxdyedydruleChainxdxdedxdxeeexyxxy yyyyyyxy1ln;ln;1;
5、1;x;eln1 )ln(ln=2005/10/7Engineering Mathematics calculus7Engineering Mathematics 微積分複習-初等函微分公式Sample: a.求出下導函 1.2.xxxfln)(=1ln1*ln*1)(+=+=xxxxxf) 1sin()(+=xxf) 1cos(1*) 1cos() 1(*) 1cos()(+=+=+=xxxdxdxxf2005/10/7Engineering Mathematics calculus8Engineering Mathematics 微積分複習-鎖規則5. Chain rule 鎖規則觀:)y
6、, x(fz =dyyzdxxzdz)()(+=ty yz tx xz tz +=2005/10/7Engineering Mathematics calculus9Engineering Mathematics 微積分複習-鎖規則Sample:) 23613() 12)(4(22*) 4( 2*) 12 (6*) 12 ( 3*) 4() 4(*) 4( 2*) 12 () 12 (*) 12 ( 3*) 4() 4(*) 12 () 12 (*) 4(:,) 12 () 4(323223322322223332322223333223322+=+=+=+=+=xxxxxxxxxxxxdxd
7、xxxdxdxxxdxdxxdxdxdxdysolxxy2005/10/7Engineering Mathematics calculus10Engineering Mathematics 微積分複習-偏導6. Partial derivative 偏導定義:推存在時,稱在對x可偏微分 稱此極限值為偏導,記為對y可偏微,同可證xyxfyxxfx+= ),(),(00000lim),(yxf),(00yx),(),(0000yxxfyxfx=2005/10/7Engineering Mathematics calculus11Engineering Mathematics 微積分複習-偏導Sam
8、ple:xy2exy)y, x(fz+=yeyxf xzxy2+=xexy2yf yzxy+=微分視為常，而對，則將如：常一變視為在做偏導時，可將另xy“ :p.s2yeyxf xzxy+=2005/10/7Engineering Mathematics calculus12Engineering Mathematics 微積分複習-初等函之積分運算2 2.對與指積分 a. b. c.d.cyndyynn+=+1 ) 1(1+=cylnydyxxdxd1)(ln=cayaydy+=)ln()(bxbxdxd =1)(ln(cedueuu+=auauaeedud=)(nnxnxdxd) 1()(
9、1+=+caadxaxx+=*ln12005/10/7Engineering Mathematics calculus13Engineering Mathematics 微積分複習-初等函之積分運算證明: b. 因為所以c. 因為所以d.因為所以ln*ln1)ln(xxx aaaaaa=caadxaxx+=*ln1xx1)(ln=+=cylnydyauauaee= )(cedueuu+=2005/10/7Engineering Mathematics calculus14Engineering Mathematics 微積分複習-初等函之積分運算Sample: 1.2.dxxxx+)ln1 (
10、4cxdxxxdxxxxxx+=+=+444*41)ln1 (4*41)ln1 (*dxexcexdedxexxx+=)(2005/10/7Engineering Mathematics calculus15Engineering Mathematics 微積分複習-初等函之積分運算3.三角函之積分 a.b.cyydy+=+12tan) 1() 1(1)(tan21 += xxdxd+=cusinuducosuududcos)(sin=+=cucosudusinuududsin)(cos=+=cuudutansec2uudud2sec)(tan=+=cuuducotcsc2uudud2csc)
11、(cot=2005/10/7Engineering Mathematics calculus16Engineering Mathematics 微積分複習-初等函之積分運算c. +=cxcoslnxdxtancxsinlnxdxcot+=cxtanxseclnxdxsec+=cxcotxcsclnxdxcsc+=2005/10/7Engineering Mathematics calculus17Engineering Mathematics 微積分複習-分部積分4.Integration by parts 分部積分公式推導:=+=+=vduuvudvvduudvuvvduudvuvd)(20
12、05/10/7Engineering Mathematics calculus18)cossin(21sin)cossin(sin2sinsincossinsinsincossin,cos,coscoscossin)cos()cos(sincos,sin,sinxxexdxexxexdxedxxexexexdxedxexxexdxexvdxeduxdxdveuxdxexdxexexdxedxexxevduuvudvxdxexvdxeduxdxdveuxdxexxxxxxxxxxxxxxxxxxxxxxx=+=+=移項並整後可得作分部積分再對求=+=+=vduuvudvvduudvuvvduudvuvd)(
