设xy^x-e^xy 2=0 求dy dx-搜答案-搜搜考试网

x2y+xy2-x-y=xy（x+y）-（x+y）=（x+y）（xy-1）∵x+y=-5，xy=7，∴原式=-5×（7-1）=-30．

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'^

(x+y)(xy)=x^2y+xy^2=-8原式=-7

siny-e^x+xy^2=0cosy.y'-e^x+2xy.y'+y^2=0(cosy+2xy)y'=e^x-y^2y'=(e^x-y^2)/(cosy+2xy)

X,Y是两个相互独立的随机变量,则D(X-Y)=D(X)+(-1)^2*D(Y)=5D(X)=E(X^2)-[E(X)]^2E(X^2)=2+1=3同理E(Y^2)=3+1=4而cov(X,Y)=0,

e^(x+y)-xy=1两边同时求导,e^(x+y)*(1+dy/dx)-y-xdy/dz=0(1)验证x=0,y=0在原曲线上.令x=0,y=0代入到（1）e^0*(1+dy/dz)-0-0*dy/

x3+y3-x2y-xy2=(x+y)(x2-xy+y2)-xy(x+y)=（x+y)(x2-2xy+y2）=（x+y)（x2+2xy+y2-4xy）=（x+y）[(x+y)2-4xy]=10×（10

由题意得，x-1=0，y+3=0，解得x=1，y=-3，所以，1-xy-xy2=1-1×（-3）-1×（-3）2，=1+3-9，=4-9，=-5．

f(x,y)=e^(x+y)+cos(xy)=0 //: 利用隐函数存在定理：f 'x(x,y)=e^

dsiny+de^x-dxy²=0cosydy+e^xdx-y²dx-2xydy=0cosydy-2xydy=y²dx-e^xdxdy/dx=(y²-e^x)/

2（xy-5xy2）-（3xy2-xy）=（2xy-10xy2）-（3xy2-xy）=2xy-10xy2-3xy2+xy=（2xy+xy）+（-3xy2-10xy2）=3xy-13xy2，∵（x+1）

解-x²y-xy²=-xy(x+y)=-2×5=-10

∵x+y=0，xy=-7，∴①x2y+xy2=xy（x+y）=-7×0=0；②x2+y2=（x+y）2-2xy=14．

方程两边同时对x求导,得y+xy'-e^x+(e^y)y'=0∴y'=(e^x-x)/(e^y+y)

是不是求：5x²y-[2x²-(3xy-xy²)-3x²]-2xy²-y²再问：是再答：已知是不是(x+3)²+|x+y+10|=

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

3xy2（x-x3y2-12x2y）=3x2y2-3x4y4-32x3y3，当xy=-1时，原式=3×（-1）2-3×（-1）4-32×（-1）3=32．

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,e^y-e^0=0,则e^y=1,则y=0所以y'(

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

/>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'