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)]
1/y = (2^x+1)/2^x = 1 + 2^(-x) >1
2^(-x) = 1/y -1
x = -log2 (1/y -1) (0
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