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求一阶偏导数:z=arctan√(x^y )
人气:482 ℃ 时间:2020-05-08 15:32:12
解答
z'(x)=1/[1+(x^y)] * 1/2√(x^y) * yx^(y-1) =yx^(y-1) / {2√(x^y)[1+(x^y)] }
z'(y)=1/[1+(x^y)] * 1/2√(x^y) * lnx *x^y=(x^y) *lnx / {2√(x^y)[1+(x^y)] }
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