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E∧x+y+Cosxy=10,求y的导数

两边求导,得到: e^x+y'-sinxy(y+xy')=0 e^x+y'-xsinxy*y'=ysinxy y'(1-xsinxy)=ysinxy-e^x y'=(ysinxy-e^x)/(1-xsinxy).

望采纳

cosxy=x -sinxy*(y+xy')=1 y+xy'=-cscxy y'=-(cscxy+y)/x. y=cos(x+y) y'=-sin(x+y)*(1+y') y'[1+sin(x+y)]=-sin(x+y) y'=-sin(x+y)/[1+sin(x+y)].

解析: 隐函数求导 cos(xy)=x -sin(xy)●(y+xy')=1 y'=[-1/sin(xy)-y]/y cos(x+y)=y -sin(x+y)●(1+y')=y' y'=-sin(x+y)/[1+sin(x+y)]

sin(x+y)-2=e^(-xy) 两边对x求导: cos(x+y)·(x+y)'-0=e^(-xy)·(-xy)' cos(x+y)·(1+y')=e^(-xy)·(-y-xy') cos(x+y)+y'·cos(x+y)=-y·e^(-xy)-xy'·e^(-xy) y'[cos(x+y)+x·e^(-xy)]=-y·e^(-xy)-cos(x+y) ∴y'=[-y·e^(-xy)-cos(x+y)]/[cos(x+y)+x·e^(...

两边同时对x求导 e^x-e^yy'=cos(xy)(y+xy') y'=[e^x-ycos(xy)]/【xcos(xy)+e^y】 dy/dx=[e^x-ycos(xy)]/【xcos(xy)+e^y】

y'=[cos(x+y)]'=-sin(x+y)*(1+y')=-sin(x+y)+-sin(x+y)*y' 把含y'的部分移到等式的右边,所以: y'=-sin(x+y)/1+sin(x+y)

要么:u(x,y) = e^x - e^y -sin(xy)...(1)..是二元函数,求偏导数; 要么:e^x - e^y -sin(xy) = 0............(2)....是隐函数,求y'。 对于(2): e^x - y'e^y - cos(xy) (y+xy') = 0 解出:y' = [e^x - ycos(xy)] / [e^y+xcos(xy)].............

Zx=∂z/∂x=ye^xy*cos(x+y)-e^xy*sin(x+y) Zy=∂z/∂y=xe^xy*cos(x+y)-e^xy*sin(x+y)

∂z/∂x =sinxy +x *cosxy *y +e^(x+y) 而 ∂z/∂y =x *cosxy *x +e^(x+y) =x^2 *cosxy +e^(x+y) 那么继续求导得到二阶导数 ∂^2z/∂x^2 =y *cosxy + y*cosxy - x *sinxy *y^2 +e^(x+y) =2y *cosxy -xy^2 *sinxy ...

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