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 $n\ge 2$

Tìm ${{y}^{(3)}}$của $y=f({{x}^{2}}+1)$

Tìm ${{y}^{(3)}}$của $y=f({{x}^{2}}+1)$
Đặt $u={{x}^{2}}+1$
$\Rightarrow {{y}^{(1)}}={{f}^{(1)}}(u).{{u}^{(1)}}(x)=2x.{{f}^{(1)}}(u)$
$\begin{align}
& {{y}^{(2)}}=2.{{f}^{(1)}}(u)+2x.2x.{{f}^{(2)}}(u) \\
& {{y}^{(3)}}=4x{{f}^{(2)}}(u)+8x.{{f}^{(2)}}(u)+2x.2x.2x.{{f}^{(3)}}(u)=12x.{{f}^{(2)}}(u)+{{(2x)}^{3}}{{f}^{(3)}}(u) \\
\end{align}$

Tìm ${{y}^{(100)}}$ của hàm số $y=\frac{2x+1}{x-1}$

Tìm ${{y}^{(100)}}$ của hàm số $y=\frac{2x+1}{x-1}$
Đặt $u=(2x+1)$ và $v={{(x-1)}^{-1}}$
Ta có ${{y}^{(n)}}=\sum\limits_{k=0}^{n}{C_{n}^{k}{{u}^{(k)}}{{v}^{(n-k)}}}$
Tính được $$\begin{align}
& C_{n}^{0}=1 \\
& C_{n}^{1}=n \\
& C_{n}^{2}=\frac{n!}{(n-2)!.2!}=\frac{1}{2}n(n-1) \\
& C_{n}^{3}=\frac{n!}{(n-3)!.3!}=\frac{1}{6}n.(n-1).(n-2) \\
\end{align}$$
$$\begin{align}
& u=(2x+1)\Rightarrow {{u}^{(0)}}=2x+1 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{u}^{(1)}}=2 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{u}^{(2)}}=0 \\
\end{align}$$
$$\begin{align}
& v={{(x-1)}^{-1}}=>{{v}^{(0)}}={{(x-1)}^{-1}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{v}^{(1)}}=(-1){{(x-1)}^{-2}}={{(-1)}^{1}}{{(x-1)}^{-(1+1)}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{v}^{(2)}}=(-1)(-2).{{(x-1)}^{-3}}=1.2.{{(x-1)}^{-3}}={{(-1)}^{2}}1.2.{{(x-1)}^{-(2+1)}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=>{{v}^{(n)}}={{(-1)}^{n}}n!{{(x-1)}^{-(n+1)}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{v}^{(n-1)}}={{(-1)}^{n-1}}(n-1)!{{(x-1)}^{-n}} \\
\end{align}$$
$$\begin{align}
& \Rightarrow {{y}^{(100)}}=(2x+1){{(-1)}^{100}}100!{{(x-1)}^{-101}}+100.2.{{(-1)}^{99}}99!{{(x-1)}^{-99}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=100!\frac{2x+1}{{{(x-1)}^{101}}}-2.100!\frac{1}{{{(x-1)}^{99}}} \\
\end{align}$$

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 $n\ge 2$
$$\begin{align}
& \Rightarrow {{y}^{(10)}}={{(-1)}^{11}}.2.4.6.8.10.\frac{1}{{{2}^{10}}}{{x}^{\frac{1}{2}-10}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-15\frac{1}{{{2}^{2}}}{{x}^{-\frac{19}{2}}} \\
\end{align}$$

Tìm đạo hàm cấp cao ${y^{(10)}}$ của hàm số $y = {x^2}{e^{2x}}$

Tìm ${{y}^{(10)}}$ của $y={{x}^{2}}{{e}^{2x}}$
Đặt$u={{x}^{2}}$
Đặt $v={{e}^{2x}}$
Ta có ${{y}^{(n)}}=\sum\limits_{k}^{n}{C_{n}^{k}}{{u}^{(n)}}{{v}^{(n-k)}}$
Lại có
$C_{n}^{0}=\frac{n!}{n!.0!}=1$
$C_{n}^{1}=\frac{n!}{(n-1)!.1!}=n$
$C_{n}^{2}=\frac{n!}{(n-2)!.2!}=\frac{1}{2}n.(n-1)$
$C_{n}^{3}=\frac{n!}{(n-3)!.3!}=\frac{1}{6}n.(n-1)(n-2)$
Ta có
$$\begin{align}
& u={{x}^{2}}\Rightarrow {{u}^{(0)}}={{x}^{2}} \\
& {{u}^{(1)}}=2x \\
& {{u}^{(2)}}=2 \\
& {{u}^{(3)}}=0 \\
\end{align}$$
Ta có
$$\begin{align}
& v={{e}^{2x}}=>{{v}^{(n)}}={{2}^{n}}{{e}^{2x}} \\
& {{v}^{(n-1)}}={{2}^{n-1}}{{e}^{2x}} \\
& {{v}^{(n-2)}}={{2}^{n-2}}{{e}^{2x}} \\
\end{align}$$
$$\begin{align}
& \Rightarrow {{y}^{(10)}}={{x}^{2}}{{2}^{10}}{{e}^{2x}}+10.2x{{.2}^{9}}.{{e}^{2x}}+\frac{1}{2}{{.10.9.2.2}^{8}}.{{e}^{2x}} \\
& ={{x}^{2}}{{2}^{10}}{{e}^{2x}}+{{10.2}^{10}}.x.{{e}^{2x}}+{{90.2}^{8}}{{e}^{2x}} \\
& ={{x}^{2}}{{2}^{10}}{{e}^{2x}}+{{5.2}^{11}}.x.{{e}^{2x}}+{{45.2}^{9}}.{{e}^{2x}} \\
\end{align}$$