Answer
a) $f''g+2f'g'+fg''$
b) $F'''=f'''g+3f''g'+3f'g''+fg''';
F^{(4)}=f^{(4)}g+4f'''g'+6f''g''+4f'g'''+fg^{(4)}$
c)$F^{(n)}=\sum_{k=0}^n {n \choose k}f^{(n-k)}g^{(k)}$
Work Step by Step
a) Applying the product rule, we see that
$$F'=f'g+fg'$$
doing it a second time, we find
$$F''=f''g+f'g'+fg''+f'g'$$ $$=f''g+2f'g'+fg''$$
b) Applying the product rule to the second derivative, we get
$$F'''=f'''g+f''g'+2(f''g'+f'g'')+f'g''+fg'''$$
$$=f'''g+3f''g'+3f'g''+fg'''$$
Doing it a second time yields
$$F^{(4)}=f^{(4)}g+f'''g'+3(f'''g'+f''g'')+3(fg''+f'g')+f'g'''+fg^{(4)}$$
$$f^{(4)}g+4f'''g'+6f''g''+4f'g'''+fg^{(4)}$$
c) Notice that each derivative follows the same pattern of 'degrees' and coefficients as powers of binomials. Because of this, we can simply use the binomial theorum as our formula for $F^{(n)}.$ The resulting formula is $$F^{(n)}=\sum_{k=0}^n {n \choose k}f^{(n-k)}g^{(k)}$$
