lncos+x-1+的导数

晁吴刘1083limx - >0 lncosx/x -
禄奋楠15966738880 ______ x->0,分子上有lncosx=cosx-1=1-x^2/2-1,所以极限为-x/2,结果为0

晁吴刘1083求导y=ln cos(2x+1) -
禄奋楠15966738880 ______[答案] y=ln cos(2x+1) y'=[ln cos(2x+1)]' =1/cos(2x+1)*[cos(2x+1)]' =1/cos(2x+1)*[-sin(2x+1)]*(2x+1)' =-2/cos(2x+1)*sin(2x+1) =-2tan(2x+1)

晁吴刘1083 求函数 y =ln(cos 2 x )的导数. -
禄奋楠15966738880 ______[答案] y′=·2cosx·(-sinx)=-2tanx.

晁吴刘1083lim x→0+ x次根号下cos 根号下x等于多少,过程 -
禄奋楠15966738880 ______ x→0+ lim (cos√x)^(1/x) =lim e^ln (cos√x)^(1/x) =e^[lim ln (cos√x)^(1/x)] 现在考虑 lim ln (cos√x)^(1/x) =lim ln(cos√x) / x =lim ln(1+(cos√x-1)) / x 利用等价无穷小:ln(1+x)~x =lim (cos√x-1) / x =-lim (1-cos√x) / x 利用等价无穷小:1-cosx~x^2/2 =-lim x/2 / x =-lim 1/2 =-1/2 因此,原极限=e^(-1/2) 有不懂欢迎追问

晁吴刘1083∫sinxcosx/1 - cos²xdx= -
禄奋楠15966738880 ______ ∫[sinxcosx/(1-cos²x)]dx=∫[cosx/(cos²x-1)]d(cosx)=½ln|cos²x-1|+C=½ln|sin²x|+C=ln|sinx| +C

晁吴刘1083∫tan√(1+x^2)xdx/√(1+x^2) -
禄奋楠15966738880 ______ 因为d√(1+x^2)=2xdx/√(1+x^2) 所以 ∫[tan√(1+x^2)]xdx/√(1+x^2)=(1/2)∫tan√(1+x^2)d√(1+x^2)=(1/2)∫[sin√(1+x^2)]/[cos√(1+x^2)]d√(1+x^2)=-(1/2)∫[1/cos√(1+x^2)]d[cos√(1+x^2)]=-(1/2)ln|cos√(1+x^2)|+C 扩展资料 不定积分的公式1、∫ a dx =...

晁吴刘1083y'=lncosx+cos1,则y＂= -
禄奋楠15966738880 ______ y'=lncosx+cos1 y''=-sinx/cosx=-tanx

晁吴刘1083ln/cos x - 1/cos x+1/如何化简为ln/csc x - cot x/ -
禄奋楠15966738880 ______[答案] ln|(cosx-1)/(cosx+1)| =ln|(1-cosx)/(1+cosx)| =ln|(1-cosx)²/[(1+cosx)(1-cosx)]| =ln|(1-cosx)²/sin²x| =2ln|(1-cosx)/sinx| =2ln|1/sinx-cosx/sinx| =2ln|cscx-cotx|

晁吴刘1083lim [ln(1+x)(√1+x - 1)]/arcsin(1 - cos x/2) -
禄奋楠15966738880 ______ lim [ln(1+x)(√1+x -1)]/arcsin(1-cos x/2)=lim [x(x/2)]/(1-cos x/2)=lim(x方/2)/ [(x/2)方/2]=(1/2)/(1/8)=4

晁吴刘1083lim x→0+ x次根号下cos 根号下x等于多少, -
禄奋楠15966738880 ______[答案] x→0+ lim (cos√x)^(1/x) =lim e^ln (cos√x)^(1/x) =e^[lim ln (cos√x)^(1/x)] 现在考虑 lim ln (cos√x)^(1/x) =lim ln(cos√x) / x =lim ln(1+(cos√x-1)) / x 利用等价无穷小:ln(1+x)~x =lim (cos√x-1) / x =-lim (1-cos√x) / x 利用等价无穷小:1-cosx~x^2/2 =-lim x/2 ...