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anonymous

  • one year ago

Please help me I give medals

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  1. anonymous
    • one year ago
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    @satellite73

  2. anonymous
    • one year ago
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  3. anonymous
    • one year ago
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    @jim_thompson5910

  4. anonymous
    • one year ago
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    @SithsAndGiggles

  5. anonymous
    • one year ago
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    @satellite73

  6. anonymous
    • one year ago
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    @LynFran could you help me?

  7. anonymous
    • one year ago
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    I already got the derivative, but I don't know how to get the answers

  8. anonymous
    • one year ago
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    |dw:1439517537638:dw|

  9. LynFran
    • one year ago
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    \[x ^{4}=\ln(xy)\] \[4x ^{3}=\frac{ 1 }{ xy }*\frac{ d }{ dx }(xy)\]\[4x ^{3}=\frac{ 1 }{ xy }*(1)(y)+(x)(1\frac{ dy }{ dx })\]\[4x ^{3}=\frac{ 1 }{ xy }*y+x \frac{ dy }{ dx }\]\[4x ^{3}=\frac{ y+x \frac{ dy }{ dx } }{ xy }\]\[4x ^{3}*xy=y+x \frac{ dy }{ dx }\]\[4x ^{3}*xy-y=x \frac{ dy }{ dx }\]\[(4x ^{4}y-y)/x=\frac{ dy }{ dx }\]... not sure..

  10. anonymous
    • one year ago
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    but you don't have xy, you should have x+y

  11. anonymous
    • one year ago
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    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

  12. LynFran
    • one year ago
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    i think i was over thinking it.. \[x ^{4}=\ln(xy)\]\[x ^{4}=lnx+lny\]\[4x ^{3}=\frac{ 1 }{ x }+\frac{ 1 }{ y }\frac{ dy }{ dx }\]

  13. anonymous
    • one year ago
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    I think you have to do the chain rule, first do the derivative of Ln and then from the inside

  14. anonymous
    • one year ago
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    like this

  15. anonymous
    • one year ago
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    |dw:1439518761802:dw|

  16. LynFran
    • one year ago
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    \[4x ^{3}=\frac{ x+y }{x y }\frac{ dy }{ dx }\]\[4x ^{3}(\frac{ xy }{ x+y })=\frac{ dy }{ dx }\]

  17. LynFran
    • one year ago
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    yes i did the chain rule..

  18. anonymous
    • one year ago
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    okay, so can we plug the values from this point?

  19. anonymous
    • one year ago
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    |dw:1439518990005:dw|

  20. anonymous
    • one year ago
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    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'

  21. LynFran
    • one year ago
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    yep so;\[4(1)^{3}(\frac{ (1)(e) }{ 1+e })=\frac{ dy }{ dx }\]

  22. anonymous
    • one year ago
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    but my choices are 3e e^4 4 1

  23. anonymous
    • one year ago
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    this is hard

  24. anonymous
    • one year ago
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    the answer is 4

  25. LynFran
    • one year ago
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    \[4x ^{3}-\frac{ 1 }{ x }=\frac{ 1 }{ y }\frac{ dy }{ dx }\]\[\frac{ 4x ^{4}-1 }{ x }=\frac{ 1 }{ y }\frac{ dy }{ dx }\]\[(y)(\frac{ 4x ^{4}-1 }{ x })=\frac{ dy }{ dx }\]\[(e)(4(1)^{4}-1)=\frac{ dy }{ dx }\]...idk

  26. anonymous
    • one year ago
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    |dw:1439519779528:dw|

  27. LynFran
    • one year ago
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    \[4x ^{3}=\frac{ x \frac{ dy }{ dx } }{ xy }+y\]\[4x ^{3}=\frac{ \frac{ dy }{ dx } }{ y }+y\]\[4x ^{3}=\frac{ \frac{ dy }{ dx } +y ^{2}}{ y }\]\[4x ^{3}y-y ^{2}=\frac{ dy }{ dx }\]\[y(4x ^{3}-y)=\frac{ dy }{ dx}\]... \[4e-e ^{2}=\frac{ dy }{ dx }\]lol getting.. 3.484

  28. LynFran
    • one year ago
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    \[4=\frac{ 1 }{ e }\frac{ dy }{ dx }+e\]\[(4-e)=\frac{ 1 }{ e }\frac{ dy }{ dx }\]\[e(4-e)=\frac{ dy }{ dx }\]

  29. LynFran
    • one year ago
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    im gonna settle for the 4...lol this one....

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