没y=y(x)是由方程xy e∧y=x 1确定的隐函数,求d²y dx²|x=0-搜答案-智慧作业帮

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'^

3x+3y=k+13x+3y=9k=8k=8x=2或x=1y=1y=2

两边对x求导：y'=(1+y')[sec(x+y)]^2得y'=[sec(x+y)]^2/{1-[sec(x+y)]^2}=1/{[cos(x+y)]^2-1}因此dy=dx/{[cos(x+y)]^

将z对x的偏导记为dz/dx,（不规范,请勿参照）(e^x)-xyz=0两边对x求导数(e^x)'-(xyz)'=0e^x-x'yz-xy(dz/dx)=0e^x-yz-xy(dz/dx)=0xy(d

不就是对x求导吗?把y看成中间变量y=y(x)说明要想导x要通过y这个中间变量两边对x求导:y^3+(3x*y^2)*dy/dx+(e^x)*siny+(e^x)*cosy*dy/dx=1/x下面你自

再答：隐函数高阶求导。再答：

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+

两边对x求导：y'e^y+(1+y')cos(x+y)=0,1）这里可得到y'=-cos(x+y)/[e^y+cos(x+y)]再对1）求导：y"e^y+(y')^2e^y+y"cos(x+y)-(1

e^y=sin(x+y)两边求导得e^y*y'=cos(x+y)(x+y)'=cos(x+y)(1+y')=cos(x+y)+y'cos(x+y)[e^y-cos(x+y)]y'=cos(x+y)y'

siny+xe^y=0确定有隐函数：y=y(x)于是,同时在两边对x求导：(siny+xe^y)'=0'y'*cosy+e^y+xy'e^y=0y'*(cosy+xe^y)=-e^yy'=-e^y/(

cos(x+y)+y=1两边同时对x求导-(1+y~)sin(x+y)+y~=0可得：=(1+y~)sin(x+y)=sin(x+y)/(1-sin(x+y))

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

方程z=xye^z两边对x求导数：∂z/∂x=ye^z+xye^z∂z/∂x∂z/∂x=ye^z/(1-xye^z)方程z=xy

y'=-2sin2(x+y)-2y'sin2(x+y)(1+2sin2(x+y))y'=-2sin2(x+y)y'=-2sin2(x+y)/(1+2sin2(x+y))

dz=-dx-dy

=-1/(4e)

分式线下的代数式请加括号,否则有歧义!再问：再问您一道题e^y(dy/dx)+1)=1再问：我用分离变量算了，就是跟答案不一样再问：您帮忙写一下详细过程再答：是否是e^y(dy/dx+1)=1?若是，

方程y=sin(x+y)两边对x求导数有：y'=cos(x+y)(x+y)'=cos(x+y)(1+y')移项整理得：[1-cos(x+y)]y'=cos(x+y)因此：y'=cos(x+y)/[1-

F(x,y)=x^2+y^2-ln(x+2y)Fx=2x-1/(x+2y)Fy=2y-2/(x+2y)F(x)=-Fx/Fy=-[2x(x+2y)-1]/[2y(x+2y)-2]