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z=y/f(x2-y2)的一阶偏导数
人气:414 ℃ 时间:2020-06-17 05:10:28
解答
ðz/ðx=-yf'(x^2-y^2)*2x/[f(x^2-y^2)]^2=-2xyf'(x^2-y^2)*/[f(x^2-y^2)]^2ðz/ðy=[f(x^2-y^2)-yf'(x^2-y^2)*(-2y)]/[f(x^2-y^2)]^2=[f(x^2-y^2)+2y^2*f'(x^2-y^2)*]/[f(x^2-y^2)]^2
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