quality:corruption
 x)

The n -th derivative of the function e 1 x {\displaystyle e^{\frac {1}{x}}} can be expressed with the Lah numbers, as follows

For example,

d d x e 1 x = − 1 x 2 ⋅ e 1 x {\displaystyle {\frac {\textrm {d}}{{\textrm {d}}x}}e^{\frac {1}{x}}=-{\frac {1}{x^{2}}}\cdot e^{\frac {1}{x}}}

d 2 d x 2 e 1 x = d d x (− 1 x 2 e 1 x) = − − 2 x 3 ⋅ e 1 x − 1 x 2 ⋅ − 1 x 2 ⋅ e 1 x = (2 x 3 + 1 x 4) ⋅ e 1 x {\displaystyle {\frac {{\textrm {d}}^{2}}{{\textrm {d}}x^{2}}}e^{\frac {1}{x}}={\frac {\textrm {d}}{{\textrm {d}}x}}\left(-{\frac {1}{x^{2}}}e^{\frac {1}{x}}\right)=-{\frac {-2}{x^{3}}}\cdot e^{\frac {1}{x}}-{\frac {1}{x^{2}}}\cdot {\frac {-1}{x^{2}}}\cdot e^{\frac {1}{x}}=\left({\frac {2}{x^{3}}}+{\frac {1}{x^{4}}}\right)\cdot e^{\frac {1}{x}}}

d 3 d x 3 e 1 x = d d x ((2 x 3 + 1 x 4) ⋅ e 1 x) = (− 6 x 4 + − 4 x 5) ⋅ e 1 x + (2 x 3 + 1 x 4) ⋅ − 1 x 2 ⋅ e 1 x = − (6 x 4 + 6 x 5 + 1 x 6) ⋅ e 1 x {\displaystyle {\frac {{\textrm {d}}^{3}}{{\textrm {d}}x^{3}}}e^{\frac {1}{x}}={\frac {\textrm {d}}{{\textrm {d}}x}}\left(\left({\frac {2}{x^{3}}}+{\frac {1}{x^{4}}}\right)\cdot e^{\frac {1}{x}}\right)=\left({\frac {-6}{x^{4}}}+{\frac {-4}{x^{5}}}\right)\cdot e^{\frac {1}{x}}+\left({\frac {2}{x^{3}}}+{\frac {1}{x^{4}}}\right)\cdot {\frac {-1}{x^{2}}}\cdot e^{\frac {1}{x}}=-\left({\frac {6}{x^{4}}}+{\frac {6}{x^{5}}}+{\frac {1}{x^{6}}}\right)\cdot e^{\frac {1}{x}}}