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z ArCtAn xy y E x

错了呵

1、本题的解答方法是:运用复合函数的链式求导法; (链式求导 = chain rule) . 2、具体解答过程如下: (若看不清楚,请点击放大,图片更加清晰) .

关键点:隐函数求导法则求偏导数

望采纳,谢谢啦

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 ...

偏导数过程如下: z=xarctanxy dz=arctanxydx+x*[1/(1+x^2y^2)]*(ydx+xdy) dz=arctanxydx+xydx/(1+x^2y^2)+x^2dy/(1+x^2y^2) 所以: z对x的偏导数为: arctanxy+xy/(1+x^2y^2) z对y的偏导数为: x^2/(1+x^2y^2)

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...

如图

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