高等数学求导

求详解
2025-03-24 18:06:38
推荐回答(2个)
回答1:

详细过程在这里,希望能帮到你,望采纳

回答2:

y= arccos[ (1-x^2)/(1+x^2)]
cosy =(1-x^2)/(1+x^2)
=-1 + 2/(1+x^2)
-siny. y' = -4x/(1+x^2)^2
y' =[4x/(1+x^2)^2] . [ 1/√ { 1- [ (1-x^2)/(1+x^2)]^2 } ]
=[4x/(1+x^2)^2] . [ (1+x^2)/(2x)]
= 2/(1+x^2)