高等数学D学习通超星期末考试答案章节答案2024年

高等数学D学习通超星期末考试章节答案2024年/star3/origin/29126bcebb21c37a88af5c4942c61088.png

答案:证:当时,与都是无穷小,且因所以,当时,与是等价无穷小。/star3/origin/a2bbf5262540f248053d353b04df6df7.png

答案:原式./star3/origin/7e63ed174129245bb337095bc504078b.png

答案:原式./star3/origin/954c304b829f61c59c4932998d61aef0.png

答案:原式./star3/origin/6e4d65c0eb8fad460249e6dd432e84d3.png

答案:原式./star3/origin/e432678cda6dbb3be7bb1cca09e9f2bd.png

答案:原式./star3/origin/52bcc80e5e61cee98a8aa6c3a70b72a5.png

答案:原式./star3/origin/c51a7ae22c59d87ba1b8ef6d2233026b.png

答案:原式./star3/origin/a93ef0e350b553168ea439e7ae034c64.png

答案:原式./star3/origin/566842189bc17347a32bedf5062c229d.png

答案:,./star3/origin/980d60584c64b26a6f2520e1a5d29dff.png

答案:得,于是,,,因为,故。/star3/origin/0dd883eee558c309b45dc5473fb81383.png

答案:;/star3/origin/e1a62af0aa9afc20097d3a1397c3617b.png

答案:(1),即,(2),.(3),所以。/star3/origin/3b13c8385a717b8cc9cea6431fcbff8e.png

答案:证明:令,一定恒不为0,于是又因为,而/star3/origin/b191b5a5624a2eae0078261892167d73.png

答案:,,所以,这样。/star3/origin/33c60b2097dd54b093880beb3ee2b5c4.png

答案:(1)因为,,;所以,在处函数连续。又因为,,;所以函数在处不可导。(2),且,所以在处连续。这个极限不存在,所以在处不可导。/star3/origin/f1ceb0a2382c1229f0782f77ac6a5b7b.png

答案:连续,且有一阶导数/star3/origin/1c359c8468e00f97b8cfdd798ef72552.png

答案:不存在/star3/origin/6c911bf5393a5602462669928e233d25.png

答案:其中所以即/star3/origin/bbe00bd4862c4a5f684bdc573912b76e.png

答案:(1)(2)(3)=xarcsinx++C/star3/origin/d829b070c151b4915579b6687279299c.png

答案:(2)(3)=/star3/origin/c78b613efcd8d71303ef5bdc0ffa9b66.png

答案:;;/star3/origin/f7c9f621b80f9f54db7688fce1a1c123.png

答案:(1)由题可得:,/star3/origin/d4f845ad367de42f210b19b04a91c80c.png

答案:令,则,,/star3/origin/17a3c60e3319eb0eebe6a954d4312c2d.png

答案:(1)原式(2)原式/star3/origin/3777f241144f9a923cf5eab455fd573c.png

答案:(1)令,则代入原式得(2)令,则代入原式得又由可得sint=x,原式=x+C/star3/origin/f2fa63606a60afe8e1452ca7d02b8e86.png

答案:,、为任意常数/star3/origin/5bb9c92a3efefbbf37a0f54ecacb42d9.png

答案:由,联立得交点.面积;体积/star3/origin/e7fceda6e1afe640e52ad22c6d519128.png

答案:(发散)/star3/origin/4a147bbd2942227e30ded42dcef07df7.png

答案:先求驻点:,解得:,形成个驻点。又由于:,可知:可得:函数在取极大值,在这四点去极小值。/star3/origin/85aa92980b121127e6b3c13749d5ac63.png

答案:设点,则:(1)时故设:,题目所求距离即为最小值可得唯一驻点及唯一极值点,即为题目所求最小值点故代入得