高数中求偏导的题目,z^3-3xyz=a^3,求d^2z/dxdy,的二次混合求导

2019-04-14

优质解答

用隐函数微分法
令F[x,y,z] = z³-3xyz-a³
z'x = -F'x/F'z = yz/(z²-xy)
z'y = -F'y/F'z = xz/(z²-xy)
(z也是y的函数,刚才我当成常数扔了- -!)
z''xy = [z'x]'y = [(yz)'(z² - xy) - yz * (2z z'y - x)]/(z²-xy)²
= [(z + y z'y)(z²-xy) - 2yz² z'y + xyz]/(z²-xy)²
= (z³ - yz² z'y - xy² z'y)/(z²-xy)²
= [z³ - (yz²+xy²)xz/(z²-xy)]/(z²-xy)²
= z(z^4 - 2xyz³ - x²y²z)/(z²-xy)³ 用隐函数微分法
令F[x,y,z] = z³-3xyz-a³
z'x = -F'x/F'z = yz/(z²-xy)
z'y = -F'y/F'z = xz/(z²-xy)
(z也是y的函数,刚才我当成常数扔了- -!)
z''xy = [z'x]'y = [(yz)'(z² - xy) - yz * (2z z'y - x)]/(z²-xy)²
= [(z + y z'y)(z²-xy) - 2yz² z'y + xyz]/(z²-xy)²
= (z³ - yz² z'y - xy² z'y)/(z²-xy)²
= [z³ - (yz²+xy²)xz/(z²-xy)]/(z²-xy)²
= z(z^4 - 2xyz³ - x²y²z)/(z²-xy)³