關(guān)于導(dǎo)數(shù)的29個(gè)典型習(xí)題

1、1關(guān)于導(dǎo)數(shù)的29個(gè)典型習(xí)題習(xí)題1設(shè)函數(shù)在的某鄰域內(nèi)類(有一階連續(xù)導(dǎo)數(shù))，且若在0?x1C.0)0(0)0(???ff)0()2()(fhfbhfa??時(shí)是比高階的無(wú)窮小，試確定的值。0?hhba解由題設(shè)知.0)0()1()]0()2()([lim0???????fbafhfbhfah由洛比達(dá)法則知.010)0(?????baf?故聯(lián)立可).0()2(1)2(2)(lim)0()2()(lim000fbahfbhfahfhbfhafhh?

2、???????????洛0)0(??f?.02??ba解出.12???ba習(xí)題2設(shè)其中有二階連續(xù)導(dǎo)數(shù)，且.(1)求(2)討論000)()(??????????xxxexgxfx)(xg1)0(1)0(????gg)(xf?在上的連續(xù)性.)(xf?)(????解(1)當(dāng)時(shí)，用公式有0?x)1()()()(])([)(22xexxgxgxxexgexgxxfxxx??????????????當(dāng)0?x時(shí)，用定義求導(dǎo)數(shù)，有.21)0()(lim

3、)0(20?????????gxexgfxx二次洛??????????????????.021)0(0)1()()()(2xgxxexxgxgxxfx(2)因在處有0?x).0(21)0(2)(lim2)1()()()(lim)(lim000fgexgxexexgxgxxgxfxxxxxx???????????????????????????洛而在處連續(xù)，故)(xf?0?x).()(??????Cxf習(xí)題3證明：若（圓），其中為定數(shù)則定

4、數(shù)。022?????cybxayxcba)04(22???cba??xdyddxdy22232])(1[證求導(dǎo)，即再導(dǎo)一次，即022??????ybayyx.22byaxy?????02222?????????ybyyy.2)1(22byyy???????3.0)(0)(.010)10(1111)()1()(00)0()()(???????????????????????????????????????ffccccfffcccfcff即

5、?從而顯然當(dāng)取.)1()(11)()(?????????????????????abacbcbcafbfa則且代入得baa???1bab????).10(1????.)()(bafbfa???????習(xí)題7求在處的100階導(dǎo)數(shù)值。)1ln()(2xxxf??0?x解由Tayl公式有.故)(98...32)(100100543xoxxxxxf??????).!97(99098!100)0(.981)0(!1001)100()100(???

6、?????ff習(xí)題8設(shè)證明2ebae???).(4lnln222abeab???證設(shè)應(yīng)用Lagrange中值定理有l(wèi)n)(2xxf?又設(shè).)(ln2lnln22baabab????????ln)(ttt??則當(dāng)時(shí)此時(shí)單減.從而ln1)(2ttt????et?0)(??t?)(t?即)()(2e????).(4lnln.2lnln222222abeabeee????????習(xí)題10設(shè)在內(nèi)有定義，存在，且滿足如果)()(xgxf)(????

7、)()(xfxf???.0)()()()(??????xfxgxfxf求證)(0)()(babfaf???.0)(bxaxf???證故使][)(baCxf??][ba????欲證).(min)()(max)(][][xfmfxfMfbaba??????只需證明反證法，若則又為極大，故.0)(bxaxf???.0??mM0?M0)()(??????fba)(?f但另一方面.0)(????f矛盾。故知若則仿上面的證明，有0)()()()()