Cho $y=\sqrt[3]{ax+b}$ tìm ${{y}^{(n)}}$

Cho $y=\sqrt[3]{ax+b}$ tìm ${{y}^{(n)}}$
Bài giải:
$y=\sqrt[3]{ax+b}\Rightarrow y={{(ax+b)}^{\frac{1}{3}}}$
$$\begin{align}
& {{y}^{(1)}}=a.\frac{1}{3}{{\left( ax+b \right)}^{\frac{1}{3}-1}} \\
& {{y}^{(2)}}={{a}^{2}}\frac{1}{3}\left( -\frac{2}{3} \right){{(ax+b)}^{\frac{1}{3}-1-1}}=(-1){{a}^{2}}\frac{1.2}{{{3}^{2}}}{{(ax+b)}^{\frac{1}{3}-1-1}} \\
& {{y}^{(3)}}=(-1).{{a}^{3}}\frac{1.2}{{{3}^{2}}}\left( -\frac{5}{3} \right){{\left( ax+b \right)}^{\frac{1}{3}-1-1-1}}={{a}^{3}}\frac{1.2.5}{{{3}^{3}}}{{\left( ax+b \right)}^{\frac{1}{3}-1-1-1}} \\
& {{y}^{(4)}}={{a}^{4}}\frac{1.2.5}{{{3}^{3}}}\left( -\frac{8}{3} \right){{\left( ax+b \right)}^{\frac{1}{3}-1-1-1-1}} \\
& {{y}^{(5)}}={{a}^{5}}\frac{1.2.5.8}{{{3}^{4}}}\left( -\frac{11}{3} \right){{\left( ax+b \right)}^{\frac{1}{3}-5}} \\
\end{align}$$
$=>{{y}^{(n)}}={{(-1)}^{n+1}}{{a}^{n}}\frac{\underset{k=2}{\overset{n}{\mathop{\prod }}}\,(3(k-1)-1)}{{{3}^{n}}}{{\left( ax+b \right)}^{\frac{1}{3}-n}}$ với $n\ge 2$