Suppose ${\text{⨍ (x)=e}}^{-\sqrt{\mathrm{x}}}{\text{+e}}^{{\mathrm{x}}^{\frac{-1}{2}}}\text{.}$  If${\text{⨍ }}^{11}\left(\mathrm{x}\right)\text{=α.}\frac{{\mathrm{e}}^{-\sqrt{\mathrm{x}}}}{\mathrm{x}}\left(1+1/\sqrt{\mathrm{x}}\right)\text{+β}\frac{{\mathrm{e}}^{{\mathrm{x}}^{\frac{-1}{2}}}}{{\mathrm{x}}^4}\left(3-\frac{2}{{\mathrm{x}}^2}\right)$ then $\left(\mathrm{\alpha }, \mathrm{\beta }\right)$

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{-\sqrt{\mathrm{x}}}+{\mathrm{e}}^{-\frac{1}{{\mathrm{x}}^2}} \\ {\mathrm{f}}^1\left(\mathrm{x}\right)={\mathrm{e}}^{-\sqrt{\mathrm{x}}}\left(-\frac{1}{2\sqrt{\mathrm{x}}}\right)+{\mathrm{e}}^{-\frac{1}{{\mathrm{x}}^2}}\left(+\frac{2}{{\mathrm{x}}^3}\right) \\ {\mathrm{f}}^{11}\left(\mathrm{x}\right)={\mathrm{e}}^{-\sqrt{\mathrm{x}}}{\left(\frac{-1}{2\sqrt{\mathrm{x}}}\right)}^2+\frac{1}{4}\frac{1}{{\mathrm{x}}^{{\displaystyle \frac{3}{2}}}}{\mathrm{e}}^{-\sqrt{\mathrm{x}}}+{\mathrm{e}}^{\frac{-1}{{\mathrm{x}}^2}}{\left(\frac{-2}{{\mathrm{x}}^3}\right)}^2-\frac{6}{{\mathrm{x}}^4}{\mathrm{e}}^{\frac{-1}{{\mathrm{x}}^2}} \\ =\frac{1}{4\mathrm{x}}{\mathrm{e}}^{-\sqrt{\mathrm{x}}}+\frac{1}{4\mathrm{x}\sqrt{\mathrm{x}}}{\mathrm{e}}^{-\sqrt{\mathrm{x}}}+{\mathrm{e}}^{\frac{-1}{{\mathrm{x}}^2}}\frac{4}{{\mathrm{x}}^6}-\frac{6}{{\mathrm{x}}^4}{\mathrm{e}}^{\frac{-1}{{\mathrm{x}}^2}} \\ =\frac{1}{4}\frac{{\mathrm{e}}^{-\sqrt{\mathrm{x}}}}{\mathrm{x}}\left(1+\frac{1}{\mathrm{x}}\right)-\frac{2}{{\mathrm{x}}^4}{\mathrm{e}}^{\frac{-1}{{\mathrm{x}}^2}}\left(3-\frac{2}{{\mathrm{x}}^2}\right) \\ \mathrm{\alpha }=\frac{1}{4}, \mathrm{\beta }=-2 \end{gathered}$$