展开全部
y′
={x′√(1-x^2)-x[√(1-x^2)]′}/(1-x^2)
={√(1-x^2)-x(1-x^2)′/[2√(1-x^2)]}/(1-x^2)
={√(1-x^2)+2x^2/[2√(1-x^2)]}/(1-x^2)
=[√(1-x^2)+x^2/√(1-x^2)]/(1-x^2)
=(1-x^2+x^2)/[(1-x^2)√(1-x^2)]
=1/[(1-x^2)√(1-x^2)]。
={x′√(1-x^2)-x[√(1-x^2)]′}/(1-x^2)
={√(1-x^2)-x(1-x^2)′/[2√(1-x^2)]}/(1-x^2)
={√(1-x^2)+2x^2/[2√(1-x^2)]}/(1-x^2)
=[√(1-x^2)+x^2/√(1-x^2)]/(1-x^2)
=(1-x^2+x^2)/[(1-x^2)√(1-x^2)]
=1/[(1-x^2)√(1-x^2)]。
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询