Question

If F(x) = f(x)g(x), where f and g have derivatives of all orders, show that F'' = f''g + 2f'g' + fg''

Answer 1

you have a product so use the product rule
\[(RS \cdot T)'=(RS)'\cdot (T)+(RS) \cdot (T)'\]
R, S, and T are functions of x
\[(R \cdot \S)'=(R)' \cdot (\S)+(R) \cdot (\S)'\]

Answer 2

i couldn't get those things to look like S's

Answer 3

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