Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 94

a.) Suppose that $F(x) = f(x) g(x)$, where $f$ and $g$ have derivatives of all orders.
$\quad$ Show that $F'' = f''g + 2f'g' + fg''$

b.) Find similar formulas $F'''$ and $F^{(4)}$
c.) Guess a formula for $F^{(n)}$


$
\begin{equation}
\begin{aligned}
\text{a.) } F(x) &= f(x)g(x)\\
F'(x) &= f'(x)g(x) + f(x)g'(x)\\
F''(x) &= [f''(x)g(x)+f'(x)g'(x)] + [f'(x)g'(x)+f(x)g''(x)]\\
F''(x) &= f''(x)g(x) + 2 f'(x)g'(x) + f(x)g''(x)\\
F''(x) &= f''g+2f'g'+fg''
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
\text{b.) } F'''(x) &= [f'''g+f''g']+2[f''g'+f'g'']+[f'g''+fg''']\\
F'''(x) &= f'''g+f''g'+2f''g'+2f'g''+f'g''+fg'''\\
F'''(x) &= f'''g+3f''g'+3f'g''+fg'''\\
F^{(4)} (x) &= [ f^{(4)}g+f'''g']+3[f'''g'+f''g'']+3[f''g''+f'g''']+[f'g'''+fg^{(4)}]\\
F^{(4)}(x) &= f^{(4)}g+f'''g'+3f'''g'+3f''g''+3f''g''+3f'g'''+f'g'''+fg^{(4)}\\
F^{(4)}(x) &= f^{(4)}g+4f'''g'+6''g''+4f'g'''+fg^{(4)}
\end{aligned}
\end{equation}
$


c.) Based on what we obtain from previous problem, $F^{(n)}$ seems to have a formula of
$\displaystyle F^{(n)} = f^{(n)}g+nf^{(n-a)}g^{(1)} + \frac{n(n-1)f^{(n-2)}g^{(2)}}{2}+ \cdots + nf^{(1)} g^{(n-1)}+fg^{(n)}$

$\displaystyle F^{(n)} = \sum\limits_{k =0}^n \frac{n!}{k!(n-l)!} f^{(n-k)} g^{(k)}$