y=cos(xy)-x的dy dx

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

x-xy=8(1)xy-y=-9(2)则有(1)-(2):X-XY-XY+Y=X+Y-2XY=8-(-9)=17

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

-sin(xy)[ydx+xdy]=2xy^2*dx+x^2*2ydy-sin(xy)ydx-sin(xy)xdy=2xy^2*dx+2x^2*ydy-2x^2*ydy-sin(xy)xdy=2xy^

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]/[

对方程两边同时求导得,﹣﹙y＋xy′﹚sin﹙xy﹚＋e^y＋﹙x＋1﹚y′e^y＝0令x=0则方程cos(xy)+(x+1)*e^y=2为1＋e^y=2,得y=0,即切点坐标为﹙0,0﹚将﹙0,0﹚

x=0时,代入原方程得：e^y-cos0=0,得：y(0)=0对x求导：e^(x+y)*(1+y')+sin(xy)*(y+xy')=0因此y'=-[ysin(xy)+e^(x+y)]/[e^(x+y

沿y＝x趋于原点时,极限为lim(1－cos(x^2+x))/2x^3趋于无穷再问：这样回答老师打了问号,是不是最后的极限不能出现x呀?再答：不是不能出现x，你可以写得再详细一点，用洛必达法则或等价替

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

xy'+y+sin(πy)πy'=0y'=-y/[x+πsin(πy)]

令y=xuy'=u+xu'代入原方程：[x(u+xu')-xu]cos²u+x=0xu'cos²u+1=0cos²udu=-dx/x(1+cos2u)du=-2dx/x积

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

x=0时,代入方程得：1+1=y,得：y=2对x求导：(y+xy')e^xy-sin(xy)*(y+xy')=y'将x=0,y=2代入得：2=y'故dy(0)=2dx

∂z/∂x=yeˆ(xy)-sin(x+y)x=1,y=0时,∂z/∂x=-sin1∂z/∂y=xeˆ(xy