cos+x+y+求导

慎朋话3518y=sin(x+y)用隐函数求导? -
支贤信13316586012 ______[答案] y=sin(x+y)等式两边同时对x求导, y'=cos(x+y)*(1+y');所以y'=cos(x+y)/[1-cos(x+y)] 注意y是x的函数!

慎朋话3518求y=sin(x+y)所确定的隐函数的二介导数 -
支贤信13316586012 ______[答案] 两边对x求导, y'=cos(x+y)*(1+y'), 再求导, y''=cos(x+y)*y''-sin(x+y))*(1+y')^2 得:y''= [sin(x+y))*(1+y')^2]/[cos(x+y)-1]

慎朋话3518sin(x+y)的导数 -
支贤信13316586012 ______ (1+y')cos(x+y)

慎朋话3518求z=xcos(x+y)的偏导数 -
支贤信13316586012 ______[答案] 对x求导时,y看成常数: z'(x)=cos(x+y)-xsin(x+y)——这用到积求导公式 对y求导时,x看成常数: z'(y)=-xsin(x+y)

慎朋话3518求sin(x+y)=sinx+siny的导数还有一题 e^x+x=e^y+y -
支贤信13316586012 ______[答案] 两边求导:cos(x+y)*(1+y')=cosx+cosy*y' y'=(cosx-cos(x+y))/(cos(x+y)-cosy) e^x+1=e^y*y'+y' y'=(e^x+1)/(e^y+1)

慎朋话3518e的y次方等于a cos(x+y)隐函数求导 -
支贤信13316586012 ______ 两边对x求导:y+xy'=e^(x+y).(1+y') 由此,解出y'即可.

慎朋话3518y=sin(x+y)求导中的1 - cos(x+y)是怎么来的 -
支贤信13316586012 ______[答案] 对x求导 则y'=cos(x+y)*(x+y)' y'=cos(x+y)*(1+y') 所以y'=cos(x+y)+y'cos(x+y) 移项就有了y'[1-cos(x+y)]=cos(x+y) y'=cos(x+y)/[1-cos(x+y)]

慎朋话3518y=sin(x+y)求导中 1 - cos(x+y) 如何得出来的 请给详细步骤 急求 -
支贤信13316586012 ______ y=sin(x+y) y'=cos(x+y)*(1+y') y'=cos(x+y)+y'cos(x+y) [1-cos(x+y)]y'=cos(x+y) y'=cos(x+y)/[1-cos(x+y)]

慎朋话3518求函数x cos y=sin(x+y)的导数 -
支贤信13316586012 ______[答案] 两边对x求导,有 cosy-xsiny*y'=cos(x+y)(1+y') cosy-cos(x+y)=y'*[cos(x+y)+xsiny] y'=[cosy-cos(x+y)]/[cos(x+y)+xsiny] 希望对你有所帮助

慎朋话3518求函数xcosy=sin(x+y)的导数 -
支贤信13316586012 ______[答案] 对x求导 1*cosy+x*(-siny)*y'=cos(x+y)*(x+y)' cosy-xsiny*y'=cos(x+y)(1+y')=cos(x+y)+cos(x+y)*y' 所以y'=[cosx-cos(x+y)]/[cos(x+y)+xsiny]