Answer:
See proof below
Step-by-step explanation:
By the rule of the derivative of a product and sum (we will omit the argument x to make clearer the calculations)
F' = (fg)' = f'g + fg'
F'' = (f'g + fg')' = (f'g)' + (fg')' = (f''g+f'g') + (f'g'+fg'') =
f''g + 2f'g' + fg''
b) In a similar way, we can find that
F''' and [tex]\bf F^{(4)}[/tex] are
F''' = f'''g + 3f''g' + 3 f'g'' + fg'''
[tex]\bf F^{(4)} = f^{(4)}g+4f'''g'+6f''g''+4f'g'''+fg^{(4)}[/tex]
c)
The pattern for higher derivatives resemble the Newton's binomial:
[tex]\bf F{(n)}=\binom{n}{0}f^{(n)}g^{(0)}+\binom{n}{1}f^{(n-1)}g^{(1)}+\binom{n}{2}f^{(n-2)}g^{(2)}+...+\binom{n}{n}f^{(0)}g^{(n)}[/tex]
where
[tex]\bf f^{(0)}[/tex] means no derivative and
[tex]\bf \binom{n}{m}[/tex] are the combination of n elements taken m at a time
[tex]\bf \binom{n}{m}=\frac{n!}{m!(n-m)!}[/tex]
