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u =x∧y z2,求一阶偏导数及全微分

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u =x∧y z2,求一阶偏导数及全微分
利用全微分的形式不变性
u =x∧y z2,求一阶偏导数及全微分
题目表达不明确!
若是 u=x^y*z^2 ,
则 u'=y*u^(y-1)z^2,u'=x^y*lnx*z^2,u'=2zx^y,
du=[yu^(y-1)z^2]dx+(x^y*lnx*z^2)dy+(2zx^y)dz.
若是 u=x^(y*z^2) ,
则 u'=yz^2*x^(y*z^2-1),u'=z^2*x^(y*z^2)lnx,u'=2yz*x^(y*z^2)lnx,
du=[yz^2*x^(y*z^2-1)]dx+[z^2*x^(y*z^2)lnx]dy+[2yz*x^(y*z^2)lnx]dz..
再问: 题目是u=x∧(y+z2)
再答: 请表达清楚啊! 若是 u=x^(y+z^2) , 则 u'=(y+z^2)x^(y*+z^2-1), u'=x^(y+z^2)lnx, u'=2z*x^(y+z^2)lnx, du=[(y+z^2)x^(y*+z^2-1)]dx+[x^(y+z^2)lnx]dy+[2z*x^(y+z^2)lnx]dz..
再问: 嗯,总之很感谢你