I'm guessing you want (fg)"' using the product rule ...
(fg)' = f'g + fg'
(fg)" = f"g + f'g' + f'g' + fg" = f"g + 2f'g' + gf'
so, what do you think?
(b) Find a similar formula for
F'''.
F''' = 3f '''g + 3f ''g' + 3f 'g'' + 3fg'''
F''' = f '''g − 3f ''g' − 3f 'g'' + fg'''
F''' = f '''g + 6f ''g'f 'g'' + fg'''
F''' = f '''g + f ''g' + f 'g'' + fg'''
F''' = f '''g + 3f ''g' + 3f 'g'' + fg'''
1 answer