若函数y=cos(3x y)的图像关于原点

cos(xy)=x+y两边分别对x求导：-sin(xy)*(y+xy′)=1+y′y′=-[1+ysin(xy)]/[xsin(xy)+1]=========左边对cos求导：-sin(xy)再对xy

两端对x求导得e^（x+y）*(x+y)'-sin（xy）*(xy)'=0e^（x+y）*(1+y')-sin（xy）*(y+y')=0解得dy/dx=y'=[e^（x+y）-ysin（xy）]/[s

我来试试吧...z=e^xy*cos(x+y)Z'x=ye^xycos(x+y)-e^xysin(x+y)Z'y=xe^xycos(x+y)-e^xysin(x+y)故dZ=[ye^xycos(x+y

functionz=yourfunc(x,y)%scriptforf(x,y)=x2+cos(xy)+2y%inputscalar:x,y%outputscalar:z%writtenbyyourna

由题知,y*(cosx+3)=cosx-3则：cosx=-3(y+1)/(y-1)由cosx的取值范围知：-1≤cosx≤1所以-1≤-3(y+1)/(y-1)≤1由-3(y+1)/(y-1)≥-1解

隐函数求导设z=x²y²-cos(xy)dy/dx=-(δz/δx)/(δz/δy)=-(2xy²+ysin(xy))/(2x²y+xsin(xy))=-y/x

cos(xy)-x^2·y=1两边对x求导-sin(xy)*(y+xy')-2xy-x^2y'=0===>x=1,y=0,y'=0-cos(xy)(y+xy')^2-(y'+y'+xy")-2y-2x

两边对x求导：-(y+xy')sin(xy)=2xy^2+2x^2yy'解得：y'=-[ysin(xy)+2xy^2]/[2x^2y+xsin(xy)]所以dy=-[ysin(xy)+2xy^2]/[

y=12[1+cos2（x-π12]+12[1-cos2（x+π12]-1=12[cos（2x-π6）-cos（2x+π6）]=sinπ6•sinx=12sinx．T=π．故答案为：π．

f(x,y)=e^(x+y)+cos(xy)=0      //: 利用隐函数存在定理：f 'x(x,y)=e^

y=cos（π/3-x）y'=-sin(π/3-x)*(-1)=sin(π/3-x)y=e^3xy'=e^(3x)*3=3e^(3x)y=In(3-x)y'=1/(3-x)*(-1)=1/(x-3)y

xy'+y+sin(πy)πy'=0y'=-y/[x+πsin(πy)]

对等式两边求导,得y'=-sin(xy)*(y+xy')y'=-ysin(xy)/[xsin(xy)+1]

应经求过导了先整体对cos求导,再对xy求导,根据乘法的求导规则就是y+xy'

画出图像即可令t=sinx所以t的范围[-1,1]y=cost[-1,1]在-π/2到π/x之间所以最大值在t=0处取得为1,最小值在t=-1或1处取得为cos1所以它的值域为1>=cos(sinx)

y'=-sin(4-3X)*(-3)=3sin(4-3X)

cos(xy)=x-y,隐函数,两边求导-sin(xy)*(xy)'=1-y'-sin(xy)*(y+xy')=1-y'-ysin(xy)-xcos(xy)*y'=1-y'y'[1-xsin(xy)]

对两边取对数：xy+3lny=lncos(x-y)两边同时对x求导：y+xy'+y'*3/y=-tan(x-y)*(1-y')整理得：y'=tan(x-y)+y/tan(x-y)-x-3/y不知道对不

复合函数求导y'=[cos（sinx）]'=sin(sinx)·（sinx）‘=sin（sinx）·cosx