Suppose that \(f(x)\) and \(g(x)\) are n-times differentiable functions. Then \(n\)-th derivative of \(f(x)g(x)\) is given by
$$\{f(x)g(x)\}^{(n)}=\frac{d^{n}}{dx^{n}}f(x)g(x)=\sum_{i=0}^{n} { }_{n}C_{i} f^{(i)}(x)g^{(n-i)}(x)$$
Notice that \({}_{n} C_{i}=\frac{n!}{i!(n-i)!}\) is the binomial coefficient and \(f^{(i)}(x)\) denotes the \(i\)th derivative of \(f\), and in particular \(f^{(0)}(x)=f(x)\).
| \(i\)th derivative | \({}_{n}\mathrm{C}_{k}\) notation | |
| \((fg)^{(0)}=fg=f^{(0)}g^{(0)}\) | \((fg)^{(0)}={}_{0}\mathrm{C}_{0}f^{(0)}g^{(0)}\) | |
| \((fg)’=f’g+fg’=f^{(1)}g^{(0)}+f^{(0)}g^{(1)}\) | \((fg)^{(1)}={}_{1}\mathrm{C}_{0}f^{(1)}g^{(0)}+{}_{1}\mathrm{C}_{1}f^{(0)}g^{(1)}\) | |
| \((fg)^{(2)}=f^{(2)}g^{(0)}+2f^{(1)}g^{(1)}+f^{(0)}g^{(2)}\) | \((fg)^{(2)}={}_{2}\mathrm{C}_{0}f^{(2)}g^{(0)}+{}_{2}\mathrm{C}_{1}f^{(1)}g^{(1)}+{}_{2}\mathrm{C}_{2}f^{(0)}g^{(2)}\) | |
| \((fg)^{(3)}=f^{(3)}g^{(0)}+3f^{(2)}g^{(1)}+3f^{(1)}g^{(2)}+f^{(0)}g^{(3)}\) | \((fg)^{(3)}={}_{3}\mathrm{C}_{0}f^{(3)}g^{(0)}+{}_{3}\mathrm{C}_{1}f^{(2)}g^{(1)}+{}_{3}\mathrm{C}_{2}f^{(1)}g^{(2)}+{}_{3}\mathrm{C}_{3}f^{(0)}g^{(3)}\) |
This rule is similar to the binomial theorem : \((x+y)^{n}=\sum_{r=0}^{n}{}_{n} C_{r} x^{r}y^{n-r}\), and can be proved easily by using the product rule and mathematical induction.
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Example 1
If \(f(x)=x^{3}\mathrm{e}^{x}\), find \(f^{(n)}(x)\).
Solution : Set \(f_{1}(x)=x^{3}\) and \(f_{2}(x)=\mathrm{e}^{x}\). Then
\begin{eqnarray*}&&f_{1}^{(n)}(x)=\mathrm{e}^{x}\\&&f^{(1)}_{2}(x)=3x^{2}, \ \ f^{(2)}_{2}(x)=6x, \ \ f_{2}^{(3)}(x)=6 \ \ f_{2}^{(n)}(x)=0\ \ (n\geq 4)\end{eqnarray*}
Using the Leibniz rule
\begin{eqnarray*}f^{(n)}(x)&=&{}_{n}C_{0}\mathrm{e}^{x}\cdot x^{3}+{}_{n}C_{1}\mathrm{e}^{x}\cdot 3x^{2}+{}_{n}C_{2}\mathrm{e}^{x}\cdot 6x+{}_{n}C_{3}\mathrm{e}^{x}\cdot 6\\&=&\mathrm{e}^{x}\{x^{3}+3nx^{2}+3n(n-1)x+n(n-1)(n-2)\}\end{eqnarray*}
Example 2
If \(f(x)=\frac{1}{x^{3}+1}\), find \(f^{(n)}(0)\).
Solution : If we let \(y=\frac{1}{x^{3}+1}\), \(y(x^{3}+1)=1\). Then by the Leibniz rule
\begin{eqnarray*}&&{}_{n}C_{0}y^{(n)}\cdot (x^{3}+1)+{}_{n}C_{1}y^{(n-1)}\cdot 3x^{2}+{}_{n}C_{2}y^{(n-2)}\cdot 6x+{}_{n}C_{3}y^{(n-3)}\cdot 6=0\\&&(x^{3}+1)y^{(n)}+3nx^{2}y^{(n-1)}+3n(n-1)xy^{(n-2)}+n(n-1)(n-2)y^{(n-3)}=0\end{eqnarray*}
If we set \(x=0\),
$$f^{(n)}(0)+n(n-1)(n-2)f^{(n-3)}(0)=0$$
and since we have
$$f(0)=1, \ \ f'(0)=f”(0)=0$$
Therefore we obtain
\begin{cases}f^{(3m)}(0)=(-1)^{m}(3m)! & \\f^{(3m+1)}(0)=f^{(3m+2)}(0)=0 & \end{cases}