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设z ArCtAn xy y E x

如上图所示。

∂z/∂x=y/(1+x²y²) ∂z/∂y=x/(1+x²y²) dz=(ydx+xdy)/(1+x²y²) 在点(1, 1), dz=(dx+dy)/2

如图

1 (x+y)/(x-y)=1+2y/(x-y) [(x+y)/(x-y)]'x=2y'/(x-y) -2y(1-y')/(x-y)^2 [(x+y)/(x-y)]'y=2/(x-y)+2y(-1)/(x-y)^2 1+(x+y)^2/(x-y)^2=2x^2+2y^2/(x-y)^2 dz=[(dy/dx)(x-y)-y(1-dy/dx)]/(x^2+y^2) *dx + (x-2y)/(x^2+y^2) dy 2 [(x+y)/(1-xy)]'x...

xy+z = arctan(x+z) 两边对 x 求偏导,y + ∂z/∂x = (1+∂z/∂x)/[1+(x+z)^2] y[1+(x+z)^2] + [1+(x+z)^2]∂z/∂x = 1+∂z/∂x ∂z/∂x = {1 - y[1+(x+z)^2]}/(x+z)^2; 两边对 y 求偏导,x ...

望采纳,谢谢啦

dz=1/[1+(xy)^2]d(xy)=(ydx+xdy)/[1+(xy)^2] 第二个如果没有写错题目的话,很难解,如果分母上sinx是平方的话,利用奇偶性,结果是0

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