Tìm ${{y}^{(10)}}$ của hàm số $y=\sqrt{x}$

Tìm ${{y}^{(10)}}$ của hàm số $y=\sqrt{x}$
Bài giải:
Ta có
$$\begin{align}
& y=\sqrt{x}={{x}^{\frac{1}{2}}} \\
& =>{{y}^{(0)}}=1.\frac{1}{{{2}^{0}}}{{x}^{\frac{1}{2}-0}} \\
& \,\,\,\,\,\,\,\,{{y}^{(1)}}=\frac{1}{2}{{x}^{\frac{1}{2}-1}}=1.\frac{1}{{{2}^{1}}}{{x}^{\frac{1}{2}-1}}={{(-1)}^{1+1}}\frac{1}{{{2}^{1}}}{{x}^{\frac{1}{2}-1}} \\
& \,\,\,\,\,\,\,\,{{y}^{(2)}}=-\frac{1}{4}{{x}^{-\frac{1}{2}-1}}=-1.\frac{1}{{{2}^{2}}}{{x}^{\frac{1}{2}-2}}={{(-1)}^{2+1}}.\frac{1}{{{2}^{2}}}{{x}^{\frac{1}{2}-2}} \\
& \,\,\,\,\,\,\,\,{{y}^{(3)}}=\left( -\frac{1}{{{2}^{2}}} \right)\left( -\frac{3}{2} \right){{x}^{-\frac{1}{2}-2}}=1.3\frac{1}{{{2}^{3}}}{{x}^{\frac{1}{2}-3}}={{(-1)}^{3+1}}.1.3\frac{1}{{{2}^{3}}}{{x}^{\frac{1}{2}-3}} \\
& \,\,\,\,\,\,\,\,{{y}^{(4)}}=\frac{3}{{{2}^{3}}}\left( -\frac{5}{2} \right){{x}^{-\frac{1}{2}-3}}={{(-1)}^{4+1}}.1.3.5.\frac{1}{{{2}^{4}}}{{x}^{\frac{1}{2}-4}} \\
& \,\,\,\,\,\,\,\,{{y}^{(5)}}=-3.5.\frac{1}{{{2}^{4}}}.\left( -\frac{7}{2} \right){{x}^{-\frac{1}{2}-4}}={{(-1)}^{5+1}}1.3.5.7.\frac{1}{{{2}^{5}}}{{x}^{\frac{1}{2}-5}} \\
\end{align}$$
$\Rightarrow {{y}^{(n)}}={{(-1)}^{n+1}}n!!.\frac{1}{{{2}^{n}}}{{x}^{\frac{1}{2}-n}}$ với $n\ge 2$
$$\begin{align}
& \Rightarrow {{y}^{(10)}}={{(-1)}^{11}}.2.4.6.8.10.\frac{1}{{{2}^{10}}}{{x}^{\frac{1}{2}-10}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-15\frac{1}{{{2}^{2}}}{{x}^{-\frac{19}{2}}} \\
\end{align}$$