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