函数 , , 由方程所确定,则 =( ).-搜答案-拍题作业网

方程两边同时求x对y的导：y+xdy/dx+1/x+2ydy/dx=0,dy/dx=-(y+1/x)/(x+2y),dy=-(y+1/x)dx/(x+2y)

y=x+lny两边同时求导得dy/dx=1+1/y*dy/dx(1-1/y)dy/dx=1dy/dx=1/(1-1/y)=y/(y-1)

方程两边求关x的导数ddx(xy)＝(y+xdydx)； ddxex+y＝ex+y(1+dydx)；所以有（y+xdy

左右对x求导有y'/y=sec²(xy)(y+xy')整理有y'=y²/(cos(xy)-xy)所以dy=(y²/(cos(xy)-xy))dx

dz=y*x^(y-1)/cosz*dx+x^y*lnx/cosz*dy

e^y-xy=ee^y·dy/dx-(y+x·dy/dx)=0e^y·dy/dx-y-x·dy/dx=0(e^y-x)·dy/dx=ydy/dx=y/(e^y-x)dy/dx不能叫做dx分之dy,因为

两边对x求导y'=1/(x+y)^2*(1+y')整理得y'=1/(x+y)^2=(coty)^2

答案是（ycosxy-1)/(1-xcosxy).亲、加油哦.

z对x的偏导xy+yz+zx=1y＋yfx'+z+xfx'＝0z对y的偏导x＋z+yfy'＋xfy'＝0z对y的偏导1+fx'+yfxy"+fy'+xfxy"=01+(fx'+fy')+(x+y)fx

这道题考查隐函数求导方法,求出x=0的倒数就是切线的斜率啦,k1=y‘,然后法线的斜率就是-1/y’.x=0代入方程,得sin0+lny=0即lny=-1解得y=1/e也就是说x=0处曲线上的点是（0

我的答案在图片里,你单击一下图片可以看得更清楚.

y'=cos(x+y)(1+y')y'=cos(x+y)/(1-cos(x+y))

=-[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(

B对方程x+cos(x+y)=0两边取微分,得dx-sin(x+y)d(x+y)=0即dx-sin(x+y)dx+sin(x+y)dy=0,整理得[1-sin(x+y)]dx=-sin(x+y0dy从

对x^2+2y^2+3z^2=18两边对x求导有:2x+6zəz/əx=0,所以əz/əx=-x/3z同理,该方程两边对y求导有:4y+6zəz/&#

y+xy'+y'/y=0//对xy和lny分别求导,注意y是x的函数y'(x+1/y)=-y//移项,合并同类项y'=-y²/(xy+1)

xy+lny=1两边求导y+xy'+y'/y=0y'=-y/(x+1/y)=-y^2/(xy+1)

我来试试吧...x²+2y²+3z²=18,两边微分2xdx+4ydy+6zdz=0dz=-x/(3z)dx-2y/(3z)dy

两边对x求导：y'=e^y+xy'e^y得：y'=e^y/(1-xe^y)再问：怎么感觉不对捏再答：是不是指数为y+1,而不是y呀？再问：指数就是y吖我题目没错再答：指数是y的话，我做的就没错。

ln(x+y)=x·lny(1+y‘)/(x+y)=lny+x/y·y‘y+y·y‘=y(x+y)lny+x(x+y)·y‘y‘=【y(x+x)lny-y】/【y-x(x+y)】再问：лл��

函数 , , 由方程 所确定,则 =( ).