两边对x求导：-sin(xy)[y+xy']=1
y+xy'=-1/sin(xy)
xy'=-y-(1/sin(xy))
y'=[-y-(1/sin(xy))]/x本回答被提问者采纳
dcos(xy)/d(xy)*d(xy)/dx=dx/dx
-sin(xy)*[y*dx/dx+x*dy/dx]=1
y+x*dy/dx=-csc(xy)
x*dy/dx=-csc(xy)-y
dy/dx=-[y+csc(xy)]/x
cos(xy)=x求隐函数的导数dy\/dx 详细
方法①：先把隐函数转化成显函数，再利用显函数求导的方法求导；方法②：隐函数左右两边对x求导（但要注意把y看作x的函数）；方法③：利用一阶微分形式不变的性质分别对x和y求导，再通过移项求得的值；方法④：把n元隐函数看作(n+1)元函数，通过多元函数的偏导数的商求得n元隐函数的导数。
cos(xy)=x-y所确定的隐函数y=y(x)的导数dy\/dx
-ysin(xy)-xcos(xy)*y'=1-y'y'[1-xsin(xy)]=1+ysin(xy)y'=[1+ysin(xy)]\/[1-xsin(xy)]也可用设二元函数f(x,y)=cos(xy)-x+y 用隐函数求导法:f'x(x,y)+f,y(x,y)*y'=0 f'x(x,y)=-sin(xy)*(xy)'-1 =-ysin(xy)-1 f'y(x,y)=-sin(xy)*(xy)'+1 =-...
求sin(xy)=x的隐函数的导数dy\/dx
cos(xy)(y+xy')=1 解出y'即得 dy\/dx=1\/xcos(xy)-y\/x
设y=y(x)是由方程cos(xy)=x 确定的隐函数,则dy是? 怎么解的
dy=-[ysin(xy)+1]dx\/[xsin(xy)]
一道大一高数题, 设方程xtany=cos(xy)确定了隐函数y=y(x),求dy\/dx.
所以由链式法则,如果有关于y的函数对x求导,最后会多出一个因子dy\/dx 左边积法则+链式,右边链式+积法则 tany+sec^2 y *dy\/dx = -sin(xy)*d(xy)\/dx tany+sec^2 y *dy\/dx = -sin(xy)*(y+x*dy\/dx)把dy\/dx的整理到一起 (sec^2 y+xsin(xy))(dy\/dx)=-ysin(xy)-tany dy\/dx=...
求导数 cos(xy)=x,
用隐函数求导法 设F(x,y)=x-cos(xy),则 F'x=1+ysin(xy),F'y=xsin(xy)所以 dy\/dx=-F'x\/F'y=-[(1+ysin(xy)]\/[xsin(xy)]
隐函数华威显函数的问题,希望前辈能详解化简过程,课本上的化解看...
没什么深奥的，记清楚一点就行：导数是关于某自变量的导数。对于隐函数，现将y当做自变量求导，然后还要乘以y的导数，即y'或dy\/dx。比如y = sin(xy)两边对x求导, 左边得y', 右边是复合函数，先视为正弦函数，得cos(xy)，然后乘以(xy)的导数: (xy)' = x'y + x(y)' = y+ xy'最后y' ...
求隐函数y的导数dy\/dx y=x^tanx
1.y=x^tanx 两边取自然对数得 lny=tanxlnx 两边对x求导得 y'\/y=sec^2xlnx+tanx\/x y'=(sec^2xlnx+tanx\/x)y=(sec^2xlnx+tanx\/x)*x^tanx 2.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(...
求下列两个隐函数的导数dy\/dx
第一个，两边对x求导有 y+xy'= e^(x+y) * (1+y')整理有 dy\/dx = y' = (e^(x+y)-y)\/(x-e^(x+y))第二个，两边对x求导有 y'= -(e^y+xy'e^y)整理有 dy\/dx = y' = -e^y\/(1+xe^y)
