在f(x-1/x)=lnx,（x>0）兩邊求導(dǎo),
[f(x-1/x)] ′=1/x
f′(x-1/x) ×(x-1/x) ′=1/x
f′(x-1/x) ×[1+(1/x²)]=1/x
f′(x-1/x)=x/(x²+1)=1/[x+(1/x)]=1/√[(x-1/x)²+2]
∴f′(x)= 1/√(x²+2),（x>0）.