设函数y=f(x),由方程xy-sin(x+y)=0,确定dy\/dx
xy-sin(x+y)=0 y+xy′-(1+y′)cos(x+y)=0 y+xy′-cos(x+y)-y′cos(x+y)=0 y′(x-cos(x+y))=cos(x+y)-y y′=[cos(x+y)-y]\/[x-cos(x+y)]
设函数y=f(x),由方程xy-sin(x+y)=0,确定dy\/dx
（cos（x+y）-y）\\(x-cos(x+y))
求帮助!!!设函数y=y(x)由方程x+y-sin(xy)=0所确定,试求dy\/dx,
2012-11-22 设y是由方程sin(xy)+ln(y-x)=x所确定的x的函... 18 2014-02-27 设方程y=sin(x+y)确定了y是x的函数,求dy\/dx。 2015-04-13 由方程ysinx-cos(x-y)=0所确定的函数的导数dy... 51 2016-07-08 设函数y=y(x)是由方程xy=e^x+y所确定的函数,求d... 2010-01-09 设y=y...
设函数y=y(x)由方程sin(x+y)+e(x方)=0所确定,则dy\/dc=
∵d（2xy）=2xyln2?d（xy）=2xyln2?（ydx+xdy） d（x+y）=dx+dy ∴2xyln2?（ydx+xdy）=dx+dy 又x=0时，y=1 ∴代入上式得：dy|x=0=（ln2-1）dx
y=f(x)由方程sin(xy)-ln(x+y)\/y=1确定,求y的导数
y' sin(xy) + xycos(xy) y' - y'\/(x+y) - y' = 1\/(x+y) - y^2 cos(xy)y' = [1\/(x+y) - y^2 cos(xy)] \/ [sin(xy) + xycos(xy) - 1\/(x+y) - 1]dy\/dx = [1 - (x+y) y^2 cos(xy)] \/ [(x+y)sin(xy) + xy(x+y)cos(xy) - x - y - ...
求解:设函数y=f(x)由方程sin(x² y)-3^y=xy所确定,求dy\/dx
解：这个题目要利用隐函数的求导法则。则sin(x^2+y)=xy （两边同时求导，还要结合复合函数的求导法则）cos（x^2+ y）*（2x+y′）=y+xy′2xcos（x^2+y）-y=xy′-y′cos（x^2+ y）2xcos（x^2+y）-y=y′（x-cos（x^2+ y））y′={2xcos（x^2+y）-y}\/（x-cos（x^2+ ...
设函数y=y(x)由方程cos(xy)-ln[(x+y)\/y]=y确定,求(dy\/dx)|{x=0}
由f(x,y)=0确定，则dy\/dx=-(df\/dx)\/(df\/dy)=-(-sin(xy)*y-y\/(x+1)\/y)\/(-sin(xy)*x-y\/(x+1)*(x+1)\/(-y^2))=(sin(xy)*y+1\/(x+1))\/(-sin(xy)*x+1\/y)，把x=0代入方程，解得y=1，代入dy\/dx，得x=0处dy\/dx=1\/1=1.dy\/dx的显式形式是不易求出的.
关于隐函数求导的问题,如图
已知 sin(x+y)=xy，求dy\/dx.解一：[cos(x+y)](1+y')=y+xy'；故得[cos(x+y)-x]y'=y-cos(x+y)；∴ y'=[y-cos(x+y)]\/[cos(x+y)-x];解二：设F(x，y)=sin(x+y)-xy=0；那么dy\/dx=-(∂F\/∂x)\/(∂F\/∂y)=-[cos(x+y)-y]\/[cos(...
求由方程xy-sin(2y²)=0所确定的隐函数y=y(x)的导数dy\/dx_百度知 ...
两边对x求导得：y+x.dy\/dx-cos(2y²).4y.dy\/dx=0 整理得:dy\/dx=y\/[4ycos(2y²)-x]
已知y=f(x)由方程sinxy-x^2=y^2所确定,则dy\/dx
sin(xy)-x^2=y^2 (x.dy\/dx +y) cos(xy)- 2x = 2ydy\/dx (2y-xcos(xy))dy\/dx = ycos(xy) -2x dy\/dx = [ycos(xy) -2x]\/(2y-xcos(xy))
