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anonymous

  • one year ago

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  1. madhu.mukherjee.946
    • one year ago
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    first find dy/dx

  2. madhu.mukherjee.946
    • one year ago
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    dy/dx=10x+4y^3d/dx

  3. anonymous
    • one year ago
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    Okay so it would be -5x/2y^3

  4. anonymous
    • one year ago
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    Oooo my bad and then would i derive that again?

  5. madhu.mukherjee.946
    • one year ago
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    yep

  6. anonymous
    • one year ago
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    Im having trouble deriving it

  7. anonymous
    • one year ago
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    Would it be10+12y^2d/dx?

  8. madhu.mukherjee.946
    • one year ago
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    wait i made a mistake

  9. madhu.mukherjee.946
    • one year ago
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    it would be dy/dx=10x+4y^3dy/dx

  10. madhu.mukherjee.946
    • one year ago
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    dy/dx-4y^3dy/dx=10x dy/dx(1-4y^3)=10x

  11. madhu.mukherjee.946
    • one year ago
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    now do second time derivative

  12. zepdrix
    • one year ago
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    `Would it be10+12y^2dy/dx?` Do you mean for your second derivative? Hmm, no. It's going to be a little more complicated than that :) You need to product rule the y^3 and y'

  13. zepdrix
    • one year ago
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    \[\large\rm 10x+4y^3y'=0\]\[\large\rm 10+\color{royalblue}{(4y^3)'}y'+4y^3\color{royalblue}{(y')'}=0\]Product rule! :)

  14. anonymous
    • one year ago
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    Okay so it would be 10 + 12y^2+4y^3?

  15. anonymous
    • one year ago
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    @zepdrix

  16. zepdrix
    • one year ago
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    \[\large\rm 10+\color{royalblue}{(4y^3)'}y'+4y^3\color{royalblue}{(y')'}=0\]\[\large\rm 10+\color{orangered}{(12y^2)}y'+4y^3\color{orangered}{(y'')}=0\]Hmm I don't see your derivative y' anywhere D:

  17. anonymous
    • one year ago
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    Would the derivative of y' just be d? That part confuses me

  18. zepdrix
    • one year ago
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    derivative of y' is y''

  19. anonymous
    • one year ago
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    5x^2+y^4=-9 diff. with respect to x \[10x+4y^3 \frac{ dy }{ dx }=0 ...(1)\] \[4y^3 \frac{ dy }{ dx }=-10x,\frac{ dy }{ dx }=\frac{ -5x }{ 2y^3 }\]diff. (1)w.r.t x again \[10+4y^3\frac{ d^2y }{ dx^2 }+12y^2\left( \frac{ dy }{ dx } \right)^2=0\] put the value of dy/dx and find \[\frac{ d^2y }{ dx^2 }\] and finally plug the values of x and y

  20. anonymous
    • one year ago
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    Oooo okay and the value of dy/dx is?

  21. anonymous
    • one year ago
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    I got 10+4d^2y/dx^2+12=0 so 26 + dy/dx=0? So it would be 26?

  22. anonymous
    • one year ago
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    @zepdrix

  23. zepdrix
    • one year ago
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    I dunno, there is too much stuff going on here :( Lemme get organized...

  24. anonymous
    • one year ago
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    Okay sorry for bothering but this is the last question I need and I really don't get it

  25. zepdrix
    • one year ago
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    If you like using the d/dx's you can do that :) But I'm gonna be lazy and use the prime notation, it's just so much "neater" to work with.\[\large\rm 5x^2+y^4=-9\]Taking first derivative, applying chain rule to the y,\[\large\rm 10x+4y^3y'=0\]Evaluate this at x=2, y=1. \[\large\rm 10(2)+4(1)^3y'=0\qquad\to\qquad \color{green}{y'=-5}\]Stick that green equation into your back pocket, save it for later. Taking second derivative, we have to apply product rule,

  26. zepdrix
    • one year ago
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    \[\large\rm 10+\color{royalblue}{(4y^3)'}y'+4y^3\color{royalblue}{(y')'}=0\]This is just me "setting up" the product rule, that's what the blue is there for. It changes to orange when I've taken the derivative,\[\large\rm 10+\color{orangered}{(12y^2y')}y'+4y^3\color{orangered}{y''}=0\]Ok with those two derivatives? :o They're a little tricky.

  27. zepdrix
    • one year ago
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    Aw mannnn, your brain esplode? :( Oh noessss, brain everywhere.. ahhh what a mess!

  28. anonymous
    • one year ago
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    Ok haha its a little confusing but I'm following you

  29. zepdrix
    • one year ago
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    So I setup the product rule between the \(\rm 4y^3\) and the \(\rm y'\) The \(\rm 4y^3\) gave us \(\rm 12y^2y'\) while the \(\rm y'\) gave us \(\rm y''\).\[\large\rm 10+(12y^2\color{green}{y'})\color{green}{y'}+4y^3y''=0\]We know that y'=-5 when x=2, y=1. Remember? We stuck it in our back pocket. Let's pull it out, along with the other information they gave us, and plug it all in. \[\large\rm 10+(12(1)^2\color{green}{(-5)})\color{green}{(-5)}+4(1)^3y''=0\]And we don't plug anything in for y'', because that's what we're trying to solve for! :)

  30. anonymous
    • one year ago
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    So 10+(12*-5)(-5+4)=0 10-60(-1)=0 50?

  31. zepdrix
    • one year ago
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    woah woah woah :O

  32. zepdrix
    • one year ago
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    That -5 shouldn't be with the 4, not sure how that happened :O

  33. anonymous
    • one year ago
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    Oops

  34. zepdrix
    • one year ago
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    Do you have to enter it into an online thing or no? :o

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