Question

If F(x) = f(x)g(x), where f and g have derivatives of all orders

Answer 1

a. By the product rule,

$F=fg\implies F'=f'g+fg'$

$\implies F''=(f''g+f'g')+(f'g'+fg'')=f''g+2f'g'+fg''$

b. By the same rule,

$F'''=(f'''g+f''g')+2(f''g'+f'g'')+(f'g''+fg''')$

$F'''=f'''g+3f''g'+3f'g''+fg'''$

and

$F^{(4)}=(f^{(4)}g+f'''g')+3(f'''g'+f''g'')+3(f''g''+f'g''')+(f'g'''+fg^{(4)})$

$F^{(4)}=f^{(4)}g+4f'''g'+6f''g''+4f'g'''+fg^{(4)}$

c. You might recognize the coefficients as those that appear in the expansion of $(a+b)^n$:

1, 1

1, 2, 1

1, 3, 3, 1

1, 4, 6, 4, 1

and so on; the general pattern (known as the general Leibniz rule) is

$F^{(n)}=\displaystyle\sum_{k=0}^n\binom nkf^{(n-k)}g^{(k)}$