y=(1+1/x)^x,
即y=e^ [x*ln(1+1/x)],
所以
y'= e^ [x*ln(1+1/x)] * [x*ln(1+1/x)] '
而
[x*ln(1+1/x)] '
= x' * ln(1+1/x) + x* [ln(1+1/x)] '
= ln(1+1/x) + x* [-(1/x^2) / (1+1/x)]
= ln(1+1/x) - 1/(x+1)
故
y'= e^ [x*ln(1+1/x)] * [x*ln(1+1/x)] '
= e^ [x*ln(1+1/x)] * [ln(1+1/x) - 1/(x+1)]
= (1+1/x)^x * [ln(1+1/x) - 1/(x+1)]