第二章導(dǎo)數(shù)與微分自測題c參考答案

1、微積分疑難問題解析與自測提高第二章導(dǎo)數(shù)與微分1自測題C參考答案一.選擇題1C2C3D4A5B6B7C8D二填空題1.23342520xy?dx??678??1!nnfx?!nx3211()()44fxfxxx??????三.計算題與證明題1求函數(shù)在處的n階導(dǎo)數(shù)。????ln1?2fxxx＝0x?????(n)0n3f?解：由萊布尼茲公式及(k為正整數(shù))，得??????????111!ln11kkkkxx?????????，???????

2、???????????????n13111!12!13!1111n2n(n)2nnn2nnnfxx2nxnnxxx?所以。??????????????????n1n31!01nn13!=2(n)nfnn2設(shè)函數(shù)在可導(dǎo)，且，若在??fx0x?????0000ff???，??????20afhbfhf??時是比h高階的無窮小，試確定ab的值。0h?解：由于，所以由于故??????0lim200hafhbfhf??????????100abf

3、，???00f，。又因??10ab??????????????001limlim0==hhafhbf2hf0afhaf2hf0hh??，????????000limlim((=0)20)hhafhf2hf2hf=affhh????????－－由此得a=2b=1.3已知是周期為5的連續(xù)函數(shù)，它在的某鄰域內(nèi)滿足關(guān)系式??fx0x?其中是時是比x高階的無窮小，且??????1sin31sin8fxfxxx????＋??x?0x?在處可導(dǎo)，求曲

4、線在點(diǎn)處的切線方程。??fx1x???y=fx????6f6解：得故??????00lim1sin31sinlim8xxfxfxxax????????????＋????131ff0－＝，微積分疑難問題解析與自測提高第二章導(dǎo)數(shù)與微分3（1）（2）??222lim21xxxfx??????021limxxfxx??解：（1）時，因此2x???2001xf???????????222.2120xxxxfxfxf????????由于因此????

5、??220lim02xfxffx??????????????????2222221limlimlim.2221200xxxxxxxfxfxff?????????????????（2）由于因此??0f=1，??????210212xxxfxfxfxxx??????????????000021021limlim2limlim0ln21ln2.xxxxxxxfxfxffxxx????????????????6設(shè)，且證明：在????fxCab?

6、????fafb0??????00.fafb????????fx內(nèi)必有一個零點(diǎn)。??ab證明：由于因此又由于????fafb0??????????limlim.xaxbfxfxfafbxaxb????????????根據(jù)函數(shù)極限的局部保號性，存在????00.fafb??????，使得11220022baba??????????????????????與，????????1200fxfxxaaxbbxaxb??????????任取。則因