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2pie

  • 2 years ago

find (dy/dx) by implicit differentiation (x/y)=(y/x)

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  1. anonymous
    • 2 years ago
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    |dw:1381786279195:dw|

  2. anonymous
    • 2 years ago
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    Quotient rule on both side

  3. 2pie
    • 2 years ago
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    |dw:1381786349930:dw|

  4. 2pie
    • 2 years ago
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    is it correct? what do I do next

  5. 2pie
    • 2 years ago
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    @iambatman

  6. myininaya
    • 2 years ago
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    Good so far.

  7. myininaya
    • 2 years ago
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    \[\frac{y-xy'}{y^2}=\frac{xy'-y}{x^2}\] I would clear the fractions first.

  8. myininaya
    • 2 years ago
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    Multiply both sides by x^2y^2.

  9. myininaya
    • 2 years ago
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    \[x^2y^2 \frac{y-xy'}{y^2}=x^2y^2\frac{xy'-y}{x^2}\] You see that we can write this without the fractional style now, right?

  10. 2pie
    • 2 years ago
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    yes by calcelling the \[y ^{2} x^{2}\] on each side respectively \[x^{2}(y-xy^1)+y^2(xy^1-y)\] then?

  11. myininaya
    • 2 years ago
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    Distribute. Put y' terms on one side and the other non-y' terms on the other side Then factor y' out. The divide by whatever is being multiplied by y'.

  12. myininaya
    • 2 years ago
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    You mean equal right not plus?

  13. 2pie
    • 2 years ago
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    yes

  14. 2pie
    • 2 years ago
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    After I multiply I got\[x^2y-x^3y^\prime =xy^2y^\prime-y^3\]

  15. myininaya
    • 2 years ago
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    Right. Continue. Put y' terms together on one side and then non-y' terms on the other.

  16. 2pie
    • 2 years ago
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    \[x^2y+y^3=xy^2y^\prime + x^3 y^\prime\]

  17. 2pie
    • 2 years ago
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    then im not sure what to do

  18. myininaya
    • 2 years ago
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    Factor the y' out on that one side.

  19. myininaya
    • 2 years ago
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    \[x^2y+y^3=y'(xy^2+x^3)\]

  20. myininaya
    • 2 years ago
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    Divide both sides by what y' is being multiplied by.

  21. myininaya
    • 2 years ago
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    You can reduce your answer.

  22. 2pie
    • 2 years ago
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    \[x^2y+y^3/xy^2+x^3\]=\[y^\prime\]

  23. myininaya
    • 2 years ago
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    \[y'=\frac{x^2y+y^3}{xy^2+x^3} =\frac{y(x^2+y^2)}{x(y^2+x^2)}\] Anything cancel?

  24. myininaya
    • 2 years ago
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    Now instead of starting the way you guys did at the beginning. I think @hartnn would like to show you a easier way.

  25. 2pie
    • 2 years ago
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    ok show me

  26. hartnn
    • 2 years ago
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    hey 2pie since we have x/y = y/x can we say x^2 = y^2 ?

  27. hartnn
    • 2 years ago
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    i just did a cross-multiplication

  28. 2pie
    • 2 years ago
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    yes

  29. myininaya
    • 2 years ago
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    I think the slope can be simplified even more by the way when you guys are done. :)

  30. hartnn
    • 2 years ago
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    and then we can differentiate x^2=y^2 to get 2x =2y dy/dx can you isolaye dy/dx from this ?

  31. 2pie
    • 2 years ago
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    yes by dividing 2y on both sides

  32. hartnn
    • 2 years ago
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    and what would u get finally ? as dy/dx ?

  33. 2pie
    • 2 years ago
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    dy/dx = x/y

  34. myininaya
    • 2 years ago
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    And recall from your initial equation y/x=x/y

  35. myininaya
    • 2 years ago
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    Now another way... x^2=y^2 or x^2-y^2=0 or (x-y)(x+y)=0 or x-y=0 or x+y=0 or y=x or y=-x If y=x, then y'=1 If y=-x, then y'=-1

  36. 2pie
    • 2 years ago
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    wow it was easier than I thought thank you very much for your help and time @myininaya @hartnn :)

  37. hartnn
    • 2 years ago
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    i see you are new here , so \[ \begin{array}l\color{red}{\text{W}}\color{orange}{\text{e}}\color{#e6e600}{\text{l}}\color{green}{\text{c}}\color{blue}{\text{o}}\color{purple}{\text{m}}\color{purple}{\text{e}}\color{red}{\text{ }}\color{orange}{\text{t}}\color{#e6e600}{\text{o}}\color{green}{\text{ }}\color{blue}{\text{O}}\color{purple}{\text{p}}\color{purple}{\text{e}}\color{red}{\text{n}}\color{orange}{\text{S}}\color{#e6e600}{\text{t}}\color{green}{\text{u}}\color{blue}{\text{d}}\color{purple}{\text{y}}\color{purple}{\text{!}}\color{red}{\text{!}}\color{orange}{\text{ }}\color{#e6e600}{\text{:}}\color{green}{\text{)}}\color{blue}{\text{}}\end{array} \]

  38. 2pie
    • 2 years ago
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    thank you Im really glad I joined

  39. myininaya
    • 2 years ago
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    Who couldn't pass up free math help?

  40. hartnn
    • 2 years ago
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    we too are glad that you joined :)

  41. 2pie
    • 2 years ago
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    thanks ;)

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