第三章導(dǎo)數(shù)與微分習(xí)題解答

1、1P61習(xí)題311、根據(jù)定義求導(dǎo)數(shù)：(1)cosyx?00000cos()coslim2sinsin22limsin()sin22lim2sin2limsin()lim22sinxxxxxxxxyxxxxxxxxxxxxxxxxx????????????????????????????????????????12(2)yx?112200012()lim()lim[]1lim1122xxxxxxyxxxxxxxxxxxxx????????

2、???????????????????(3)yxx?033022302202()lim()lim[()]33lim[()]33lim()3232xxxxxxxxxxyxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx??????????????????????????????????????????????????(4)xya?001limlimxxxxxxxaaayaxx??????????????設(shè)，則tx?

3、?01limtxtayat????再設(shè)，則，于是tsa?logats?111111011limlog1limlog1limlog[1(1)]1loglnxsaxssaxssaxaxsyasasasaeaa???????????????2、30000()()()lim()()lim()()lim[()]()lim()xxxxfxxfxfxxfxxfxxfxxfxxfxxfxxfx??????????????????????????????

4、???????????所以為奇函數(shù)。()fx?（3）設(shè)為可導(dǎo)的周期函數(shù)且其周期為()fx，則：T00()()()lim()()lim()xxfxTxfxTfxTxfxxfxxfx????????????????????所以仍為以為周期的周期函數(shù)。()fx?T5、解：00||1xxxye??????在點處的切線斜率為1，法線xye??(01)斜率為了1，故所求的切線和法線方程分別為：11yxyx?????即。11yxyx?????6、解：

5、001lim()limsin0(0)xxfxxfx??????所以在處連續(xù)；()fx0x?000000011()sinsinlim()lim111sinsinsinlim11sinsin1limlimsin11(sinsin)1limsin11112cossin12sinlimxxxxxxxxxxxxxfxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx??????????????????????????????????

6、?????????????????????????0022sin112()sin2coslim1sin112()2()sin2coslim2()111sin2cos()2111sincosxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx?????????????????????????????????即：111()cossinfxxxx???于是在處不可導(dǎo)。()fx0x?7、解：在處連續(xù)，()fx?1x?112lim()