方程y=sin(x y)确定了y是x的函数,求dy dx
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/22 09:59:55
实际上先有一个微分dπy^2这里把πy^2看做一个以y为变量的函数f(y)欲求dπy^2/dx(这里有一个前提是导数是可以看做微分之商的)分母分子同乘dy,变为(dπy^2/dy)*(dy/dx)这时
sin(xy)+In(y-x)=x两边同时对x求导得cos(xy)·(xy)'+1/(y-x)·(y-x)'=1cos(xy)·(y+xy')+1/(y-x)·(y'-1)=1①当x=0时,sin0+
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)·(
y=sin(x+y),y'=cos(x+y)*(1+y'),y'=cos(x+y)/(1-cos(x+y))=dy/dx
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))
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
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
dy=dsin(x+y)dy=cos(x+y)d(x+y)dy=cos(x+y)(dx+dy)dy=cos(x+y)dx+cos(x+y)dy所以dy/dx=cos(x+y)/[1-cos(x+y)]
Fx=e^x-y^2Fy=cosy-2xydy/dx=-Fx/Fy=(y^2-e^x)/(cosy-2xy)
将x=0代入方程得:lny=1,得y=e方程两边对x求导:y+xy'+e^xlny+y'e^x/y=0代入x=0,y=e得:e+lne+y'/e=0,得y'=-e(e+1)即y'(0)=-e(e+1)
∵siny+e^x-xy^2=0,∴(dy/dx)cosy+e^x-[y^2+2xy(dy/dx)]=0,∴(cosy-2xy)(dy/dx)=y^2-e^x,∴dy/dx=(y^2-e^x)/(co
(cos(x+y)-y)\(x-cos(x+y))
dy/dx=-fx/fy,你自己可以算吧
e^(x+y)+sin(xy)=1e^(x+y)*(1+y')+cos(xy)(y+xy')=0y'*[e*(x+y)+xcos(xy)]=-[ycos(xy)+e^(x+y)]y'=-[ycos(x
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)]=[