arctan(I +t)dt 2 li 四.(本题满分7分) 设函数u=f(x,y,)有连续偏导数,且z=x(x,y)由方程xe-ye=ze所确定,求da 解法1设F(x,y,z)=xe’-ye-e',则 F,=(x+1)e (y+1)e F,=-(x+1)e 故 x+1 dz f, +f. f, +f f,+,=,-f 所以 x + adx+,-f 2十 解法2在xe-ye'=ze'两边微分,得 dx xe'dx-e'dy-ye'dy edz zedz 故 k=(1+ a)edr-(1 +y)edy (1+z)e