y是由方程sin(xy)-1 (y-x)=1所确定的函数,求y
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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)
sin(xy)+ln(y-x)=x两边同时对x求导得:cos(xy)·(y+xy')+(y'-1)/(y-x)=1①当x=0时,sin0-lny=0,解得y=1把x=0,y=1代入①得:cos0·(1
xy+e^y=y+1(1)求d^2y/dx^2在x=0处的值:(1)两边分别对x求导:y+xy'+e^yy'=y'y/y'+x+e^y=1(2)(2)两边对x再求导一次:(y'y'-yy'')/y'^
1)y|x=o当x=0时sin(0)-1/y-0=1得:y|x=0=-1(2)y'|x=osin(xy)-1/y-x=1两边对x求导:cos(xy)(y+xy')+y'/y^2-1=0当x=0时y=-
是把y看作关于x的函数.再问:不是很懂,给个步骤吧。谢谢。再答:1/y-x是(1/y)-x的意思,还是1/(y-x)?再问:1/(y-x)再答:把y看做x的复合函数,两边对x求导,得cos(xy)·(
答案是(ycosxy-1)/(1-xcosxy).亲、加油哦.
两边求导得:cos(xy)*(y+xy')+1+y'=0y'[xcos(xy)+1]=-ycos(xy)-1所以,y'=-[ycos(xy)+1]/[xcos(xy)+1]
再答:隐函数高阶求导。再答:
e^(xy)+sin(xy)=y(y+xy')e^(xy)+(y+xy')cos(xy)=y'y'=(ye^(xy)+ycos(xy))/(1-xe^(xy)-xcos(xy))
原方程是xy=1-e^y?如果是的话将等式两边对X求导数得y+xy'=e^y*y'则y‘=y/(e^y-x)y'(0)=y/e^y
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
两边对x求导得y+xy'=(1+y')/(x+y)y(x+y)+x(x+y)y'=1+y'y'[x(x+y)-1]=1-y(x+y)y'=[1-y(x+y)]/[x(x+y)-1]dy=[1-y(x+
将原方程两边微分得d[xe^y+sin(xy)]=0→e^ydx+xe^ydy+cos(xy)(ydx+xdy)=0→移项[xe^y+xcos(xy)]dy=-[e^y+ycos(xy)]dx整理→d
e^x-e^y=sin(xy)e^x-e^y*y'=cos(xy)*(y+xy')y'=(e^x-ycos(xy))/(e^y+xcos(xy))dy=(e^x-ycos(xy))/(e^y+xcos
这是隐函数.二阶导再导一次就是.方程两边对x求导,得z'=cos(xz)(xz)'+y(y不是关于x的函数吧?)=zcos(xz)+xz'cos(xz)+y所以z'=[zcos(xz)+y]/[1-x
在方程中令x=0可得,0=lney(0)+1,从而可得,y(0)=e2将方程两边对x求导数,得:cos(xy)(y+xy′)=1x+e−y′y将x=0,y(0)=e2代入,有e2=1e−y′(0)e2
(cos(x+y)-y)\(x-cos(x+y))
dy/dx=-fx/fy,你自己可以算吧
化为:e^(ylnx)-e^y=sin(xy)两边对x求导:e^(ylnx)(y'lnx+y/x)-y'e^y=cos(xy)(y+xy')y'[lnxe^(ylnx)-e^y-xcos(xy)]=[