y=ln(x+1/x)
y'=(x+1/x)'/(x+1/x)
=(1-1/x²)/(x+1/x)
=(x²-1)/(x³+x)
如图所示
{ln[x+(1/x)]}'=1/[x+(1/x)]×[x+(1/x)]'=1/[x+(1/x)]×[1-(1/x²)]=(x²-1)/(x³+x)
