Z=y⼀(x^2+y^2)^1⼀2对x的偏导怎么求,对y的偏导呢?

2024-12-23 12:30:35
推荐回答(1个)
回答1:

对x求偏导时,y是常数,变量只有分母有,那么
z=y(x^2+y^2)^(-1/2),所以əz/əx=y(x^2+y^2)^(-3/2) *(-1/2) *2x
=-xy/(x^2+y^2)^(3/2)
对y求偏导时,x是常数,分子和分母都有变量那么
əz/əy=(x^2+y^2)^(-1/2)+y(x^2+y^2)^(-3/2) *(-1/2) *2y
=1/(x^2+y^2)^(1/2)-y^2/(x^2+y^2)^(3/2)
=x^2/(x^2+y^2)^(3/2)