已知 f(x)=ln[x+√(1+x²)]；求 xf'(x)=?解：f'(x)=[x+√(1+x²)]'/[x+√(1+x²)]=[1+x/√(1+x²)]/[x+√(1+x²)]=[√(1+x²)+x]/[x√(1+x²)+1+x²]=[x+√(1+x²)]/{[x+√(1+x²)]•√(1+x²)}=1/√(1+x²);∴xf'(x)=x/√(1+x²);
