y=ln[x+(1+x^2)^(1/2)],
dy/dx={1/[x+(1+x^2)^(1/2)]}*【x+(1+x^2)^(1/2)】‘
={1/[x+(1+x^2)^(1/2)]}*【1+x/(1+x^2)^(1/2)】
=1/(1+x^2)^(1/2)
所以
dx/dy=(1+x^2)^(1/2)[x+(1+x^2)^(1/2)]'怎么会等于这个呢 1+x/(1+x^2)^(1/2)[x+(1+x^2)^(1/2)]'=1+[1/2(1+x^2)^(1/2)]×（1+x²）’ = 1+x/(1+x^2)^(1/2) =【x+(1+x^2)^(1/2)】/(1+x^2)^(1/2)