设y=yx是由方程y^3+xy+x^2-2x+1=0所确定并且满足-搜答案-神马作文网

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)

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'^

左右对x求导有y'/y=sec²(xy)(y+xy')整理有y'=y²/(cos(xy)-xy)所以dy=(y²/(cos(xy)-xy))dx

如图所示,最后求解是自上而下带入的

若x+y=4，xy=3，求yx

∵x+y=4，xy=3，∴原式=x2+y2xy=(x+y)2−2xyxy=16−63=103．

这是一个复合函数求导,y=y(x)所以求e^y的导数首先对整体求导,再对y求导即为e^y*y'xy的导数为y+x*y'(根据求导规则)所以两边求导可得e^y*y'-y-x*y'=0

两边对x求导有y'e^y=y+xy'整理解得y‘=dy/dx=x/(e^y-x)

e^z-z+xy^3=0偏z/偏x：z'e^z-z'+y^3=0y^3=z'(1-e^z)z'=y^3/(1-e^z)偏z/偏y:z'e^z-z'+3xy^2=0z'=3xy^2/(1-e^z)偏z/

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))

网上有很多高数课后习题答案,你可以下载一个参考~e^y-e^x=xy两边求导,得e^y*y'-e^x=y+xy'(e^y-x)y'=(e^x+y)所以y'=(e^x+y)/(e^y-x)x=0时,原式

xy+e^y=1e^y(0)=1y(0)=0xy'+y+e^yy'=00+y(0)+y'(0)=0y'(0)=0xy''+y'+y'+e^yy''+(y')^2e^y=00+2y'(0)+y''(0)

xy+yx=10x,y是多少

xy+yx=10x+y+10y+x=11x+11y=100+x10x=100-11yx=10-1.1y所以y只能是0

∵x-y=4xy，∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112．故答案为：112．

89

dy/dx=(1+y^2)/[xy(1+x^2)]y/(1+y^2)dy=dx/[x(1+x^2)]2y/(1+y^2)dy=2xdx[x^2(1+x^2)]d(y^2)/(1+y^2)=d(x^2)

x=0时,代入方程得：1+1=y,得：y=2对x求导：(y+xy')e^xy-sin(xy)*(y+xy')=y'将x=0,y=2代入得：2=y'故dy(0)=2dx

两边对x求导数,得y'*e^y+y+xy'=0,在原方程中令x=0可得y=1,因此,将x=0,y=1代入上式可得y'+1=0,即y'(0)=-1.再问：对x求导时y可以当成一个常数吗？为什么要用公式（

/>e^y+xy+e^x=0两边同时对x求导得：e^y·y'+y+xy'+e^x=0得y'=-(y+e^x)/(x+e^y)y''=-[(y'+e^x)(x+e^y)-(y+e^x)(1+e^y·y'

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)]=[