cos(x y)=4 5

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

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]/[

∂Z/∂x=y*cos(xy)-2cos(xy)*sin(xy)*y=y*cos(xy)-y*sin(2xy)∂Z/∂y=x*cos(xy)-2cos(

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

Zx=ycos(xy)-2ycos(xy)sin(xy)=ycos(xy)-ysin(2xy)Zy=xcos(xy)-xsin(2xy)

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'

cos(xy)=x两边对x求导：-sin(xy)[y+xy']=1y+xy'=-1/sin(xy)xy'=-y-(1/sin(xy))y'=[-y-(1/sin(xy))]/x

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

对两边分别求导,得dy/dx=-sin(xy)*(x*dy/dx+y)则dy/dx(1+sin(xy)*x)=-sin(xy)*y所以dy/dx=(-sin(xy)*y)/(1+sin(xy)*x)

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

在方程ex+y+cos（xy）=0左右两边同时对x求导，得：ex+y（1+y′）-sin（xy）•（y+xy′）=0，化简求得：y′＝dydx＝ysin(xy)−ex+yex+y−xsin(xy)．