If F(x) = f(x)g(x), where f and g have derivatives of all orders (a) Show that F ′′ = f ′′g + 2f ′ g ′ + fg′′ . (b) Find similar formulas for F ′′′ and F (4) . (c) Describe the pattern for higher derivatives of F.
Accepted Solution
A:
a. By the product rule,[tex]F=fg\implies F'=f'g+fg'[/tex][tex]\implies F''=(f''g+f'g')+(f'g'+fg'')=f''g+2f'g'+fg''[/tex]b. By the same rule,[tex]F'''=(f'''g+f''g')+2(f''g'+f'g'')+(f'g''+fg''')[/tex][tex]F'''=f'''g+3f''g'+3f'g''+fg'''[/tex]and[tex]F^{(4)}=(f^{(4)}g+f'''g')+3(f'''g'+f''g'')+3(f''g''+f'g''')+(f'g'''+fg^{(4)})[/tex][tex]F^{(4)}=f^{(4)}g+4f'''g'+6f''g''+4f'g'''+fg^{(4)}[/tex]c. You might recognize the coefficients as those that appear in the expansion of [tex](a+b)^n[/tex]:1, 11, 2, 11, 3, 3, 11, 4, 6, 4, 1and so on; the general pattern (known as the general Leibniz rule) is[tex]F^{(n)}=\displaystyle\sum_{k=0}^n\binom nkf^{(n-k)}g^{(k)}[/tex]