y=cos(xy)-x,则dy=

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

(1+X4)COSx先求导数再乘以dx就行了

最好的办法是求对数：lny=xlncosx,两边求导数得：y'/y=lncosx-x(sinx/cosx)=lncosx-xtanx所以：y'=y(lncosx-xtanx)=(cosx)^x(lnc

解析2xdx+ydx+xdy+3y²dy=0(2x+y)dx+(x+3y²)dy=0(2x+y)dx=-(x+3y²)dydy/dx=(2x+y)/-(x+3y²

dcos(xy)=dx-sin(xy)d(xy)=dx-sin(xy)(ydx+xdy)=dx-ysin(xy)dx-xsin(xy)dy=dxdy=-[ysin(xy)+1]dx/[xsin(xy)

x^2*dy/dx=xy-y^2dy/dx=y/x-y^2/x^2u=y/xy=xuy'=u+xu'代入：u+xu'=u+u^2xu'=u^2du/u^2=dx/x-1/u=lnx+lnCCx=e^(

方法一（微分法）d(y/x)=d(ln(xy))(xdy-ydx)/x²=1/xy*d(xy)即（xdy-ydx)/x²=(ydx+xdy)/xy∴dy/dx=(xy+y²

对dy=cos(1+x)d(1+x)=cos(1+x)dx

两边求导(y'x-y)/x^2=(y+xy')/xyxy+x^2y'=xyy'+y^2y'=(xy-y^2)/(xy+x^2)

两边对x求导有y'e^y=y+xy'整理解得y‘=dy/dx=x/(e^y-x)

=-[ysin(xy)+2e^(2x+y)]/[ysin(xy)+e^(2x+y)]*(dx)再问：麻烦给我写出解的过程。。再答：等式两边取对数，得：d[e^(2x+y)]-d[cos(xy)]=0(

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

解析dy/dx=-3sin(1-3x)所以dy=3sin(1-3x)dx

隐函数直接求导数.e^(x+y)(1+y')-sin(xy)(y+xy')=0解出y'即为dy/dx=[e^(x+y)-y*sin(xy)]/[e^(x+y)-x*sin(xy)]

我算的结果和你的一样,也是y'=sin(x+y)/1-sin(x+y)应该是书上写错了.在说xsin(x+y)中的x从何而来?找不到它的来源啊.不管是对cos(),还是对y求导都不会出现xsin()这

你提的问题,涉及到书本上一个重要的知识点——复合函数的求导和倒数形式的不变性.建议你回到书本扎实基础!此处的dx-dy=d（x-y）,即对（x-y）求导.举个简单的例子,d（cosx）=sinxdx,

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)]

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

y'=3[cos(1/x)]^2*[cos(1/x)]'..=3[cos(1/x)]^2*[-sin(1/x)]*(1/x)'..=3[cos(1/x)]^2*[-sin(1/x)]*(-1/x^2)

∵x=cos(y/x)==>1=-sin(y/x)*(xy'-y)/x^2(等式两端对x求导)==>xy'-y=-x^2/sin(y/x)==>xy'=y-x^2/sin(y/x)==>y'=y/x-