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Mathematics 7 Online
OpenStudy (anonymous):

If f(3)=−3, g(3)=9, f′(3)=4, and g′(3)=−1, find the following numbers: {f/(f-g)}' (3) = ______

OpenStudy (anonymous):

we are given all the values...I just dont think I should use the quotient rule for this question. The question before this was [f/g]' (3) = ___ and I used the quotient rule for that and got the correct answer. But, the denominator for this question has stumped me. I don't know what to do with the f-g in the denominator...quotient rule doesnt make sense.

OpenStudy (bibby):

just input 3-9 and proceed?

OpenStudy (anonymous):

that doesnt work

OpenStudy (gorv):

\[\frac{ d(\frac{ f }{ f-g }) }{ dx}=(f-g)*\frac{ d f }{ dx }-\frac{ f*d(f-g) }{ dx }\]

OpenStudy (bibby):

how is it formatted? \(\huge [{\frac{f}{(f-g)}}]'(3)\)?

OpenStudy (gorv):

this will be your numerator

OpenStudy (gorv):

\[(f-g)^2\]

OpenStudy (gorv):

is your denominator

OpenStudy (anonymous):

yea it's formatted that way. let me check gorv's work..

OpenStudy (gorv):

quotient rule

OpenStudy (gorv):

as wedont know the f what it is so df/dx=f '

OpenStudy (gorv):

or in place of x you can take any other variable

OpenStudy (gorv):

f on differentiated will be written as f '

OpenStudy (anonymous):

oh wow that's confusing

OpenStudy (gorv):

first numerator (f-g) * f ' - f( f ' - g ')

OpenStudy (gorv):

which one???

OpenStudy (anonymous):

the numerator

OpenStudy (gorv):

you know quoteint rule???

OpenStudy (bibby):

he's just applying the quotient rule \(\huge \frac{d}{dx}\frac{u}{v}=\frac{u'v-uv'}{v^2}\)

OpenStudy (gorv):

we dont know the variable or f is function of x , y, z etc

OpenStudy (gorv):

see @bibby posted the quoteint rule

OpenStudy (anonymous):

oh I get it!!! you used (f'-g') as being g'(x)

OpenStudy (gorv):

g' (x) = g' i used

OpenStudy (anonymous):

and g(x) is just (f-g)

OpenStudy (gorv):

can u do it now??

OpenStudy (gorv):

no no no

OpenStudy (gorv):

that we have to calculate f/(f-g)

OpenStudy (gorv):

f and g are different

OpenStudy (anonymous):

I know the quotient rule...i'm trying to figure out what values you placed for g(x)f'(x) - ....

OpenStudy (gorv):

ohhhhhh wait

OpenStudy (bibby):

I think by "g(x)=(f-g) they just meant the denominator is being treated as a term itself

OpenStudy (gorv):

f(x)/g(x)

OpenStudy (bibby):

yeah

OpenStudy (gorv):

if you considering this than g(x)= f-g

OpenStudy (gorv):

yeah u r right now just solve it

OpenStudy (gorv):

and tell us what u got ...so we can leave happily :P

OpenStudy (anonymous):

yes!! ! okay im going to solve it! and ill let you know if I get it correct. it's all online

OpenStudy (anonymous):

hahaha ok! give me a min

OpenStudy (gorv):

yeah waitinnnnnn

OpenStudy (bibby):

I can't leave happily with this depression :'(

OpenStudy (anonymous):

wehooooo!!!!!! got it right. -33/144

OpenStudy (anonymous):

you ok bibby?

OpenStudy (anonymous):

Thanks gorv and bibby!

OpenStudy (gorv):

okkk ...bibby what happened????????

OpenStudy (bibby):

Yeah, it was a joke :D good job

OpenStudy (anonymous):

your explanations were great!

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