anonymous
  • anonymous
Please help me I give medals
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
anonymous
  • anonymous
anonymous
  • anonymous

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anonymous
  • anonymous
anonymous
  • anonymous
anonymous
  • anonymous
@LynFran could you help me?
anonymous
  • anonymous
I already got the derivative, but I don't know how to get the answers
anonymous
  • anonymous
|dw:1439517537638:dw|
LynFran
  • LynFran
\[x ^{4}=\ln(xy)\] \[4x ^{3}=\frac{ 1 }{ xy }*\frac{ d }{ dx }(xy)\]\[4x ^{3}=\frac{ 1 }{ xy }*(1)(y)+(x)(1\frac{ dy }{ dx })\]\[4x ^{3}=\frac{ 1 }{ xy }*y+x \frac{ dy }{ dx }\]\[4x ^{3}=\frac{ y+x \frac{ dy }{ dx } }{ xy }\]\[4x ^{3}*xy=y+x \frac{ dy }{ dx }\]\[4x ^{3}*xy-y=x \frac{ dy }{ dx }\]\[(4x ^{4}y-y)/x=\frac{ dy }{ dx }\]... not sure..
anonymous
  • anonymous
but you don't have xy, you should have x+y
anonymous
  • anonymous
I did it like 5 times an Im sure I did it correctly you don't have to show me all the work, I know how to get to the answer, I just can find how that point value
LynFran
  • LynFran
i think i was over thinking it.. \[x ^{4}=\ln(xy)\]\[x ^{4}=lnx+lny\]\[4x ^{3}=\frac{ 1 }{ x }+\frac{ 1 }{ y }\frac{ dy }{ dx }\]
anonymous
  • anonymous
I think you have to do the chain rule, first do the derivative of Ln and then from the inside
anonymous
  • anonymous
like this
anonymous
  • anonymous
|dw:1439518761802:dw|
LynFran
  • LynFran
\[4x ^{3}=\frac{ x+y }{x y }\frac{ dy }{ dx }\]\[4x ^{3}(\frac{ xy }{ x+y })=\frac{ dy }{ dx }\]
LynFran
  • LynFran
yes i did the chain rule..
anonymous
  • anonymous
okay, so can we plug the values from this point?
anonymous
  • anonymous
|dw:1439518990005:dw|
anonymous
  • anonymous
that's what I got, I think that's wrong to be honest because if you did the chin rule, you should have found the derivative of the inside, which would be 1+y'
LynFran
  • LynFran
yep so;\[4(1)^{3}(\frac{ (1)(e) }{ 1+e })=\frac{ dy }{ dx }\]
anonymous
  • anonymous
but my choices are 3e e^4 4 1
anonymous
  • anonymous
this is hard
anonymous
  • anonymous
the answer is 4
LynFran
  • LynFran
\[4x ^{3}-\frac{ 1 }{ x }=\frac{ 1 }{ y }\frac{ dy }{ dx }\]\[\frac{ 4x ^{4}-1 }{ x }=\frac{ 1 }{ y }\frac{ dy }{ dx }\]\[(y)(\frac{ 4x ^{4}-1 }{ x })=\frac{ dy }{ dx }\]\[(e)(4(1)^{4}-1)=\frac{ dy }{ dx }\]...idk
anonymous
  • anonymous
|dw:1439519779528:dw|
LynFran
  • LynFran
\[4x ^{3}=\frac{ x \frac{ dy }{ dx } }{ xy }+y\]\[4x ^{3}=\frac{ \frac{ dy }{ dx } }{ y }+y\]\[4x ^{3}=\frac{ \frac{ dy }{ dx } +y ^{2}}{ y }\]\[4x ^{3}y-y ^{2}=\frac{ dy }{ dx }\]\[y(4x ^{3}-y)=\frac{ dy }{ dx}\]... \[4e-e ^{2}=\frac{ dy }{ dx }\]lol getting.. 3.484
LynFran
  • LynFran
\[4=\frac{ 1 }{ e }\frac{ dy }{ dx }+e\]\[(4-e)=\frac{ 1 }{ e }\frac{ dy }{ dx }\]\[e(4-e)=\frac{ dy }{ dx }\]
LynFran
  • LynFran
im gonna settle for the 4...lol this one....

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