求方程y=sin(x y)-搜答案-神马作文网

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)·(

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))

满意请采纳,谢谢

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

设Y=y'降阶：Y'=(Y/x)ln(Y/x)这就是一个一阶齐次方程.设Y/x=u,所以Y=ux,Y'=u+x(du/dx),代回原方程,解得：lnu=C1x+1Y=xe^(C1x+1)所以y=[(C

cos(x+y)(1+y')=y+xy'dy/dx=y'=[y-cos(x+y)]/[cos(x+y)-x]

详细答案在下面.希望对你有所帮助!

Fx=e^x-y^2Fy=cosy-2xydy/dx=-Fx/Fy=(y^2-e^x)/(cosy-2xy)

y+xy'-cos(πy²)2πyy'=0y=[2πycos(πy²)-x]y'y'=y/[2πycos(πy²)-x]即：dy/dx=y/[2πycos(πy²

∵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

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

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)]=[