Question

设u=f（√（x^2+y^2）,z）,其中f具有二阶连续偏导数...

Answer 1

xy+x+y-z=e^z 两端对x求偏导：
y+1-z'x=z'xe^z z'x=(y+1)/(1+e^z)
两端对y求偏导：
x+1-z'y=z'ye^z z'y=(x+1)/(1+e^z)
u'x=f'1*1/2(x^2+y^2)^(-1/2)*2x+f'2*(y+1)/(1+e^z)
=xf'1(x^2+y^2)^(-1/2)+f'2*(y+1)/(1+e^z)
u''xy=(u'x)'y
=-xyf'1(x^2+y^2)^(-3/2)+x(x^2+y^2)^(-1/2)[f''11*1/2(x^2+y^2)^(-1/2)*2y+f''12*(x+1)/(1+e^z)]
+f'2[1+e^z-(y+1)e^z(x+1)/(1+e^z)]+(y+1)/(1+e^z)[f''21*1/2(x^2+y^2)^(-1/2)*2y+f''22*(x+1)/(1+e^z)]
=-xyf'1(x^2+y^2)^(-3/2)+x(x^2+y^2)^(-1/2)[f''11*y(x^2+y^2)^(-1/2)+f''12*(x+1)/(1+e^z)]
+f'2[1+e^z-(y+1)e^z(x+1)/(1+e^z)]+(y+1)/(1+e^z)[f''21*y(x^2+y^2)^(-1/2)+f''22*(x+1)/(1+e^z)]
∵f具有二阶连续偏导数.∴f''12=f''21 下面由你自己整理.