Find a formula for (fg)''(d) = f''(d)g(d) + 2f'(d)g'(d) f(d)g''(d).

Question

anonymous · Answer

there is another plus sign after (d) + f(d)g''(d)

anonymous · Answer

what formula is it going to look like i'm confused

freckles · Answer

what are you finding a formula for?
what does that mean?
do you mean you are trying to solve the differential equation?

anonymous · Answer

i'm supposed to take the derivative of the product rule to find the formula i suppose

freckles · Answer

well (fg)''=f''g+2g'f'+g''f as you have above

anonymous · Answer

when i look in the book for an example all it gave was bionomial formula but i don't understand

freckles · Answer

are you asking why is it true?

anonymous · Answer

need a formula that says it equals that

anonymous · Answer

looks like Leibniz rule

freckles · Answer

(fg)'=f'g+fg'
(fg)'' <--to find this you have to use product rule twice 
once for the f'g and another time for the fg'
(fg)''=(f'g)'+(fg')'
     =f''g+f'g' + f'g'+fg''
     =f''g+2f'g'+fg''
so the formula you found for (fg)'' is correct

anonymous · Answer

thanks for your help. I just thought there would be more to it

freckles · Answer

like I still don't know what your question really is lol

freckles · Answer

Are you trying to say you found (fg)'' to be f''g+2g'f'+g''f
and you actually want to find (fg)^(n) ?

anonymous · Answer

yeah i think so

freckles · Answer

you think so?

freckles · Answer

is there anyway you can give me the full question?

anonymous · Answer

hard to know with no examples lol

anonymous · Answer

Find a formula for (fg)''(d) (the second derivative of the product of f and g), by taking the derivative of the product rule ( and using the product rule to do so).

freckles · Answer

oh yeah if that is the question 
then you did do that

anonymous · Answer

ok thanks for your help:) just thought there was way more to it. I have two more similar problems but i'll work on those first

freckles · Answer

just for fun we can find (fg)'''
and we can see something rather pretty 
like a pattern in the answers
$$(fg)'''=([fg]'')'=(f''g+2f'g'+fg'')' \ =f'''g+f''g'+2f'g''+2f''g'+f'g''+fg''' \ =f'''g+3f''g'+3f'g''+fg'''$$
look at this
what does this remind of you 
the answer you had before and this one right here?

freckles · Answer

|dw:1425837315919:dw|
do you know what this is?