在f(x-1/x)=lnx,(x>0)两边求导,[f(x-1/x)] ′=1/xf′(x-1/x) ×(x-1/x) ′=1/xf′(x-1/x) ×[1+(1/x²)]=1/xf′(x-1/x)=x/(x²+1)=1/[x+(1/x)]=1/√[(x-1/x)²+2]∴f′(x)= 1/√(x²+2),(x>0).
令W=x-1/x 则x=1/2(w±√(w^2+4))去掉负根代入原式得f(w)=ln(1/2(w+√(w^2+4)))然后再对f(w)求导
正解