1v微分计算题试卷1．设xyx5sincose，求yd．．解：因为)(coscos5)(sine4sinxxxyxxxxxsincos5cose4sin所以xxxxyxd)sincos5cose(d4sin2．计算积分e1dlnxxx．．解：e12e12e1)d(ln21ln2dlnxxxxxxx414ed212e2e12xx3．设xxyxcose，求yd．．解：212cos23cos23)sin(e)()(cosexxxxyxxxxxyxd)esin23(d2cos214．．计算积分xxxd1sin2．．解：cxxxxxx1cos)1(d1sind1sin25．.设xyxtanesin，求yd．.解：由导数运算法则和复合函数求导法则得)tane(ddsinxyx)(tand)e(dsinxxxxxxdcos1)(sinde2sinxxxxxdcos1dcose2sinxxxx)dcos1cose(2sin6．.计算xxxde．解：由不定积分的凑微分法得)(de2dexxxxxcxe27．.已知2sin2xx，求y．.解：由导数运算法则和复合函数求导法则得)(sin2sin)2()sin2(222xxxyxxx)(cos2sin2ln2222xxxxx22cos22sin2ln2xxxxx8．计算xxxdcos22π0．解：由定积分的分部积分法得xxxxxxxd2sinsin2dcos22π02π02π0作业（1）xxxye，求yxxxxxxxxxxxye)1(21)e(e)(21)e()(（2）bxyaxsine，求yd解：)(cose)(esin)(sine)e(sinbxbxaxbxbxbxyaxaxaxaxbxbbxaaxaxcosesinedxbxbbxdxadxydyax)cossin(e2（3）xxyx1e，求yd解：xxxdxydyxd)e123(12xxxxxxxxy12211231e12323)1(e)()e(（4）2ecosxxy，求ydxxxxxxxyxxx2sine2)(e)(sin)e()(cos2222xxxxdxydyxd)2sine2(2（5）)1ln(2xxy，求y222222112211)1(11)1(xxxxxxxxxxxy2222111)1(1xxxxxx（6）xxxd22Cxxdxxdx232221222)2(31)2()2(2