z＝y/f（x2-y2）的一阶偏导数

Question

如题所述

Answer 1

ðz/ðx=-yf'(x^2-y^2)*2x/[f(x^2-y^2)]^2=-2xyf'(x^2-y^2)*/[f(x^2-y^2)]^2

ðz/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