问题及解答

设 $f,g$ 均为 $3$ 阶可导函数, 求复合函数 $f(g)$ 的各阶导数.

\[
\begin{aligned}
(f(g))'&=f'(g)g',\\
[f(g)]''&=[f'(g)g']'=[f'(g)]'g'+f'(g)g''=f''(g)g'g'+f'(g)g'',
\end{aligned}
\]
从而
\[
\begin{split}
[f(g)]^{(3)}&=[f''(g)g'g'+f'(g)g'']'\\
&=f'''(g)(g')^3+f''(g)2g'g''+f''(g)g'g''+f'(g)g'''\\
&=f^{(3)}(g)(g')^3+3f''(g)g'g''+f'(g)g^{(3)}.
\end{split}
\]

经过计算

\[
[f(g)]^{(4)}=f^{(4)}(g)(g')^4+6f^{(3)}(g')^2g''+3f''(g)(g'')^2+4f''(g)g'g^{(3)}+f'(g)g^{(4)},
\]

\[
\begin{split}
[f(g)]^{(5)}&=f^{(5)}(g)(g')^5+10f^{(4)}(g)(g')^3 g''+10f^{(3)}(g)(g')^2 g^{(3)}+15f^{(3)}(g) g'(g'')^2\\
&\quad+10f''(g) g'' g^{(3)}+5f''(g) g' g^{(4)}+f'(g)g^{(5)}.
\end{split}
\]

\[
\begin{split}
[f(g)]^{(4)}&=\bigl[f^{(3)}(g)(g')^3+3f''(g)g'g''+f'(g)g^{(3)}\bigr]'\\
&=[f^{(4)}(g)(g')^4+f^{(3)}(g)3(g')^2 g'']+3[f^{(3)}(g)g'g'g''+f''(g)g''g''+f''(g)g'g^{(3)}]\\
&\qquad+[f''(g)g'g^{(3)}+f'(g)g^{(4)}]\\
&=f^{(4)}(g)(g')^4+6f^{(3)}(g)(g')^2g''+3f''(g)(g'')^2+4f''(g)g'g^{(3)}+f'(g)g^{(4)}
\end{split}
\]

\[
\begin{split}
[f(g)]^{(5)}&=\bigl[f^{(4)}(g)(g')^4+6f^{(3)}(g)(g')^2g''+3f''(g)(g'')^2+4f''(g)g'g^{(3)}+f'(g)g^{(4)}\bigr]'\\
&=\bigl(f^{(5)}(g)(g')^5+f^{(4)}(g)4(g')^3 g''\bigr)+6\Bigl[f^{(4)}(g)(g')^3 g''+f^{(3)}(g)g''' (g')^2+f^{(3)}(g)g'' 2g' g''\Bigr]\\
&\quad+3\Big[f'''(g)g' (g'')^2+f''(g)2g'' g'''\Bigr]+4\Bigl[f'''(g)(g')^2 g^{(3)}+f''(g)g'' g^{(3)}+f''(g)g' g^{(4)}\Bigr]\\
&\quad+\bigl(f''(g)g' g^{(4)}+f'(g)g^{(5)}\bigr)\\
&=f^{(5)}(g)(g')^5+10f^{(4)}(g)(g')^3 g''+10f^{(3)}(g)(g')^2 g^{(3)}+15f^{(3)}(g) g'(g'')^2\\
&\quad+10f''(g) g'' g^{(3)}+5f''(g) g' g^{(4)}+f'(g)g^{(5)}.
\end{split}
\]