设xy^x-e^xy 2=0 求dy dx
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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'