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y E xy的导数

两边对x求导dy/dx=e^(xy)(y+x dy/dx)dy/dx=ye^(xy)/[1-xe^(xy)]

解:该题为隐函数求导.xy+e^(xy)=1则 y+xy'+e^(xy)( y+xy')=0解得:y'=-y/x解答完毕.

两边求导,就是1+y'=(y+xy')e^xy 且x+y=e^xy,化简得到y'=(xy+y^2-1)/(1-x^2-xy)=(ye^xy-1)/(1-xe^xy)

若:e^(xy) = c ----- (0)问题为隐函数求导两边对x求导:e^(xy) (y+xy') = 0y+xy' = 0y' = -y/x ---------------------- (1)xy = ln c ------------------------(2)y = lnc / x -----------------------(3)y' = - lnc / x ---------------------(4)实际上,由(2)解出:y = lnc/x ---------------------------(5

y=e^(xy) 两边求导 dy/dx=[e^(xy)](y+x*dy/dx) 移项 dy/dx=ye^(xy)/(1-x)

隐函数求导 一次全导,y'*e^y+xy'+y=0 => y'=-y/(e^y+x) 两边再取全导 y''*e^y+(y')^2*e^y+xy''+y'+y'=0 (e^y+x)*y''+e^y*(y')^2+2y'=0 x=0, y(0)=1, y'(0)=-e^(-1), e*y''(0)+e*e^(-2)+2[-e^(-1)]=0 ey''(0)=-e^(-1)+2e^(-1)=e^(-1) =1/e y''(0)=1/e^2

e^y=xy,求y的导数 解一:两边取对数得 y=lnx+lny,两边对x取导数得 y′=1/x+y′/y(1-1/y)y′=1/x,∴y′=y/[x(y-1)] =y/(e^y-x) 解二:两边对x取导数:(e^y)y′=y+xy′(e^y-x)y′=y,故y′=y/(e^y-x)

求全导:左边=e^xy(ydx+xdy),右边为0 左边等于右边 所以dy=-y/xdx

y*e^(xy)dx +x*e^(xy)dy

y=xy,方程两边求导.y'=x'y+xy' y'=y+xy' y'(1-x)=y y'=y/(1-x)

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