设f(x,y,z)=ex+y2z,其中z=z(x,y)是由方程x+y+z+xyz=0所确定的隐函数,则fx’(0,1,一1)=____________.

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问题 设f(x,y,z)=ex+y2z,其中z=z(x,y)是由方程x+y+z+xyz=0所确定的隐函数,则fx’(0,1,一1)=____________.

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答案1

解析 已知f(x,y,z)=ex+y2z,那么有fx’(x,y,z)=ex+y2zx’.在等式x+y+z+xyz=0两端对x求偏导可得1+zx’+yz+xyzx’=0.由x=0,y=1,z=一1,可得zx’=0.故fe’(0,1,一1)=e0=1.
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