2020-12-04

## alculus – $n$th derivative of $e^{1/x}$

The Question : 61 people think this question is useful I am trying to find the $n$’th derivative of $f(x)=e^{1/x}$. When looking at the first few derivatives I noticed a pattern and eventually found the following formula $$\frac{\mathrm d^n}{\mathrm dx^n}f(x)=(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 n+k}$$ I tested it for the first