Question

一道关于高等数学偏导数的题目，望大神解答！感激不尽！

Answer 1

f=e^(x+y) *[x^1/3*(y-1)^1/3 +y^1/3*(x-1)^2/3]
于是对x求偏导数得到
f 'x=e^(x+y) *[x^1/3*(y-1)^1/3 +y^1/3*(x-1)^2/3]
+ e^(x+y) *[1/3 *x^(-2/3)*(y-1)^1/3 +2/3*y^1/3*(x-1)^(-1/3) ]
而f 'y=e^(x+y) *[x^1/3*(y-1)^1/3 +y^1/3*(x-1)^2/3]
+ e^(x+y) *[1/3 *x^1/3*(y-1)^(-2/3) +1/3*y^(-2/3)*(x-1)^2/3 ]
代入x=0，y=1
所以得到
f 'x(0,1)=e +e *(-2/3)=e/3
f 'y(0,1)=e + e *(1/3)=4e/3
选择C，就是你要的结果