(a)- If F(x) = f(x)g(x), where f and g have derivatives of all orders, show that F"=f''g+2f'g'+fg'' (b)- Find simmiliar formula for F''' and F^(4) (c)-Describe the pattern for higher derivatives of F.

Answer:See proof belowStep-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]