2.1導(dǎo)數(shù)概念

1、第2章導(dǎo)數(shù)與微分21導(dǎo)數(shù)概念習(xí)題解11用定義求函數(shù)在處的導(dǎo)數(shù)。2yx?1x?【解法一】利用導(dǎo)數(shù)定義式求解：0000()()()=lim??????xfxxfxfxx0(1)(1)(1)lim???????xfxffx220(1)1lim???????xxx。202lim???????xxxx0lim(2)2??????xx【解法二】利用導(dǎo)數(shù)定義式求解：0000()()()=limhfxhfxfxh???。0(1)(1)(1)limhfh

2、ffh????220(1)1limhhh????202limhhhh???0lim(2)2hh????【解法三】利用導(dǎo)數(shù)定義式求解：0000()()()=limxxfxfxfxxx???。1()(1)(1)lim1xfxffx????2211lim1xxx????1(1)(1)lim1xxxx?????1lim(1)2xx????2已知一物體的運(yùn)動(dòng)方程為（），求該物體在（）時(shí)的瞬時(shí)速度。38st??m2t?s【解】該物體在（）時(shí)的瞬時(shí)速

3、度為：2t?s2()(2)(2)lim2????tstsst332(8)(28)lim2??????ttt328lim2????ttt（）。22(2)(24)lim2??????ttttt22lim(24)????ttt2222412?????ms3求在拋物線上點(diǎn)處的切線方程與法線方程。22yx??1x?【解】由于，知在拋物線上點(diǎn)處的切線斜率為，2yx?22yx??1x?1122xxyx????拋物線上在點(diǎn)處切點(diǎn)的縱坐標(biāo)為，22yx??

4、1x?211(2)3xxyx?????于是，得所求的切線方程為，亦即；32(1)yx???210xy???法線方程為，亦即。13(1)2yx????270xy???（課本后的答案中，法線斜率錯(cuò)為）12第2章導(dǎo)數(shù)與微分21導(dǎo)數(shù)概念習(xí)題解3整理得，0000()(2)lim3()xfxxfxxfxx?????????即得。0000()(2)3lim()22xfxxfxxfxx?????????5用導(dǎo)數(shù)的定義求在點(diǎn)處的導(dǎo)數(shù)。0()ln(1)0x

5、xfxxx???????0x?【解】由于是函數(shù)的分段點(diǎn)，而分段點(diǎn)兩側(cè)的函數(shù)表達(dá)式不同，故須應(yīng)用左、右導(dǎo)0x?數(shù)進(jìn)行分析：利用導(dǎo)數(shù)定義式，且，0000()()()=limxxfxfxfxxx???0(0)=ln(1)ln10????xfx得，0()(0)(0)=lim0xfxffx?????00lim0xxx?????0lim11x????，0()(0)(0)=lim0xfxffx?????0ln(1)0lim0xxx??????10li

6、mln(1)xxx????ln1e??由于，(0)(0)1ff????知在點(diǎn)處的導(dǎo)數(shù)為。0()ln(1)0xxfxxx???????0x?(0)1f?6試討論函數(shù)在處的連續(xù)性與可導(dǎo)性。21sin0()00xxfxxx????????0x?【解】由于，0()(0)(0)lim0xfxffx????201sin0lim0xxxx????01limsinxxx??0?知該函數(shù)在處可導(dǎo)，0x?而函數(shù)可導(dǎo)必連續(xù)，從而知該函數(shù)在處連續(xù)可導(dǎo)。0x?7