I need to show that derivative of $\frac{d}{dn}(1+\frac{\epsilon}{2n})^n > 0.$
I use formula $(a^x)' = a^x\ln x.$
For now i have: $\frac{d}{dn}(1+\frac{\epsilon}{2n})^n = (1+\frac{\epsilon}{2n})^n \ln{(1+\frac{\epsilon}{2n})}(-\frac{\epsilon}{2n^2}),$ where $-\frac{\epsilon}{2n^2}$ is derivative of $(1+\frac{\epsilon}{2n}).$ But this is less than zero. What is the error?
On
Note: $a^x = (e^{\ln a})^x = e^{x \ln a}$, so by the Chain Rule $(a^x)' = e^{x \ln a}(\ln a) = a^x \ln a$.
(Thus the formula you quoted has a typo.)
More importantly, though, that formula only applies if the base $a$ is a constant; in your case the base is itself a function of $n$, the variable of differentiation.
If your variable appears in the base only (and the exponent is a constant), then you use the Power Rule. If your variable appears in the exponent only (and the base is a constant), then you use the formula above. However, if your variable appears in both the base and the exponent, then you should probably use Logarithmic Differentiation.
Hint. You may write $$ (1+\frac{\epsilon}{2n})^n=e^{n \log\left( 1+ \frac{\epsilon}{2n}\right)} $$ and use $$ (e^u)'=u'\times e^u $$ giving here $$ \left(n \log\left( 1+ \frac{\epsilon}{2n}\right) \right)'\times e^{n \log\left( 1+ \frac{\epsilon}{2n}\right)} $$ or $$ \left(\log\left( 1+ \frac{\epsilon}{2n}\right)- \frac{\epsilon}{2n}\frac{1}{\left(1+ \frac{\epsilon}{2n}\right)} \right)\times(1+\frac{\epsilon}{2n})^n. $$