设函数z=x^2+y^2 ,而y=y(x)由方程(e^xy)-y=0所确定,求az/ax

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设函数z=x^2+y^2 ,而y=y(x)由方程(e^xy)-y=0所确定,求az/ax
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设函数z=x^2+y^2 ,而y=y(x)由方程(e^xy)-y=0所确定,求az/ax
设函数z=x^2+y^2 ,而y=y(x)由方程(e^xy)-y=0所确定,求az/ax

设函数z=x^2+y^2 ,而y=y(x)由方程(e^xy)-y=0所确定,求az/ax
(e^xy)-y=0两端对x求导得
e^(xy)(y+xy')-y'=0
y'=y*e^(xy)/[1-xe^(xy)]
函数z=x^2+y^2
∂z/∂x=2x+∂z/∂/y*dy/dx
=2x+2y*y*e^(xy)/[1-xe^(xy)]
=2x+2y^2*e^(xy)/[1-xe^(xy)]

z=x^2+y^2 => dz = 2x dx + 2y dy ①
e^(xy) - y =0 => e^(xy) * ( y dx + x dy) - dy = 0, 即 y^2 dx = ( 1-xy) dy ②
由 ② , 得: dy = y^2 dx / (1-xy)
代入① 中: dz = 2x dx + 2y^3 /(1-xy) dx
=> dz/dx = 2x + 2y^3/(1-xy)