y=ln(x+√(1+x^2))的导数

y=ln(x+√(1+x^2))
y'=1/[x+√(1+x^2)] *[x+√(1+x^2)]'
又∵ [x+√(1+x^2)]'=1+(1/2)(1+x²)^(-1/2)*2x=1-x*(1+x²)^(-1/2)*=1-x/√(1+x^2)
∴ y'=1/[x+√(1+x^2)] * [1-x/√(1+x^2)]
=1/√(1+x^2)*{[x+√(1+x^2)]*[√(1+x^2)-x]}
=1/√(1+x^2)1+(1/2)(1+x²)^(-1/2)*2x=1-x*(1+x²)^(-1/2)*前面不是”1+“么...为啥后面就”1-“了抱歉，写错了y=ln(x+√(1+x^2))y'=1/[x+√(1+x^2)]*[x+√(1+x^2)]'又∵ [x+√(1+x^2)]'=1+(1/2)(1+x²)^(-1/2)*2x=1+x*(1+x²)^(-1/2)*=1+x/√(1+x^2)∴ y'=1/[x+√(1+x^2)] * [1+x/√(1+x^2)]={1/[x+√(1+x^2)]}* [√(1+x^2)+x]/[√(1+x^2)}=1/√(1+x^2)[√(1+x^2)+x]/[√(1+x^2)}这一步怎么来的？通分即可