www.gsyw.net > 设y=(x)由方程E平方是xy+y的3次方5x=0所确定,试求Dx分之Dy|x=0。

设y=(x)由方程E平方是xy+y的3次方5x=0所确定,试求Dx分之Dy|x=0。

e^xy +y^3 -5x=0那么对x求导得到e^xy *(y+xy') +3y^2 *y' -5=0即化简得到y'=(5-y*e^xy) / (x*e^xy+3y^2)而x=0时,y= -1代入解得dy/dx |x=0 = (5+1)/3 =2

(1)求导(ye^xy+xy'e^xy)+(3y^2*y')-(5)=0y'=(5-ye^ey)/(3y^2+xe^ey)(2)两边求导1=[1+y']/(x+y)y'=x+y-1

用隐函数求导法,得到 exy(yxy)3y2 y-5=0.其中的y'就是所求的dy/dx.把x=0带入原方程得到y=-1,然后把x=0,y=-1,带入方程得到y'=dy/dx=2.

x=0,得1+y^3=0y=-1两边对x求导,得e^(xy) (y+xy')+3yy'-5=0x=0,y=-1代入,得1*(-1)+3y'(0)-5=03y'(0)=6y'(0)=2

dy=(-y/e的y次方+x)dx

这是隐函数x=0时,代入方程得:e^y=e,得y(0)=1方程两边对x求导:y'e^y+y+xy'=0,得y'=-y/(e^y+x)x=0时,y'(0)=-1/e再对y'求导:y"=-[y'(e^y+x)-y(y'e^y+1)]/(e^y+x)代入x=0,y(0)=1,y'(0)=-1/e,得y"(0)=-[-1/e*e-(

∵e^xy-x^2+y^3=0,∴(e^xy+e^xdy/dx)-2x+3y^2dy/dx=0,∴(3y^2+e^x)dy/dx=2x-e^xy,∴dy/dx=(2x-e^xy)/(3y^2+e^x).

首先当x=0时,y=1.等式两边对x求导:y′e^y+y+xy′=0,所以y′=-y/(x+e^y) y″=y[2(x+e^y)-ye^y]/(x+e^y) 所以y″(0)=e/e=1/e

e^y-xy=ee^ydy/dx-(y+xdy/dx)=0e^ydy/dx-y-xdy/dx=0(e^y-x)dy/dx=ydy/dx=y/(e^y-x)dy/dx不能叫做dx分之dy,因为这是个导数符号,而不是分数!

x^3+y^3=e^xy 对x求导3x+3y*y'=e^(xy)*(xy)'3x+3y*y'=e^(xy)*(y+x*y')3x+3y*y'=e^(xy)*y+e^(xy)*x*y'y'=[3x-e^(xy)*y]/[e^(xy)*x-3y]即dy/dx=[3x-e^(xy)*y]/[e^(xy)*x-3y]x=0,代入x^3+y^3=e^xy 0+y^3=1y=1代入dy/dx=[3x-e^(xy)*y]/[e^(xy)*x-3y]所以dy/dx=(0-1)/(0-3)=1/3

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