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mathcalculus Group Title

Help! Implicit Differentiation with Radicals :(

  • one year ago
  • one year ago

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  1. mathcalculus Group Title
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    • one year ago
  2. Loser66 Group Title
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    what do you get so far?

    • one year ago
  3. mathcalculus Group Title
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    I'll send it right now, 1 sec.

    • one year ago
  4. mathcalculus Group Title
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    • one year ago
    1 Attachment
  5. mathcalculus Group Title
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    @Loser66

    • one year ago
  6. Loser66 Group Title
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    me here but don't know the reason why cannot open the attachment

    • one year ago
  7. Loser66 Group Title
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    take derivative both sides, the right hand side give you 0, the left hand side has 2 terms, what do you get for the first term

    • one year ago
  8. mathcalculus Group Title
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    i got some of it but i didn't get (3+2y) in the numerator.

    • one year ago
  9. mathcalculus Group Title
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    |dw:1381795463590:dw|

    • one year ago
  10. Loser66 Group Title
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    \[(\sqrt{3x+2y})' +(\sqrt{4xy})'= \frac {3+2y'}{2\sqrt{3x+2y}} +\frac{4y+4xy'}{2\sqrt{4xy}}\]

    • one year ago
  11. Loser66 Group Title
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    you have to use chain rule,

    • one year ago
  12. Loser66 Group Title
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    |dw:1381795702483:dw|

    • one year ago
  13. Loser66 Group Title
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    right?

    • one year ago
  14. mathcalculus Group Title
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    |dw:1381795570646:dw|

    • one year ago
  15. Loser66 Group Title
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    the net is sooooo bad, ok, let me write it out.

    • one year ago
  16. mathcalculus Group Title
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    so why did you bring this on top in the numerator?

    • one year ago
  17. Loser66 Group Title
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    • one year ago
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  18. Loser66 Group Title
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    y is function of x, cannot take derivative as x, derivative of y is y'

    • one year ago
  19. Loser66 Group Title
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    ridiculous the net, me here but I am invisible, hehehe

    • one year ago
  20. phi Group Title
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    |dw:1381798106573:dw|

    • one year ago
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