是y=ln[x√(1+x²)]?
y'=[x√(1+x²)]'/[x√(1+x²)]
={√(1+x²)+x*(1+x²)'/[2√(1+x²)]}/[x√(1+x²)]
=[√(1+x²)+x²/√(1+x²)]/[x√(1+x²)]
=[(1+x²)+x²]/√(1+x²)/[x√(1+x²)]
=(1+2x²)/[x(1+x²)]
=1/x+x/(1+x²)