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ksaimouli

dy/dx

  • one year ago
  • one year ago

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  1. ksaimouli
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    |dw:1350749050460:dw|

    • one year ago
  2. Chlorophyll
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    Just product rule term by term! Start with u = x² -> u' = ...?

    • one year ago
  3. ksaimouli
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    |dw:1350749192642:dw|

    • one year ago
  4. ksaimouli
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    ?

    • one year ago
  5. ksaimouli
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    power rule and then e^x is e^x

    • one year ago
  6. ksaimouli
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    i am not sure

    • one year ago
  7. ksaimouli
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    @Chlorophyll

    • one year ago
  8. Chlorophyll
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    If you're unsure, so please don't jump ahead! You have your choices: following my guidance by fill in the blanks or show your work completely :)

    • one year ago
  9. ksaimouli
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    |dw:1350749473100:dw|

    • one year ago
  10. ksaimouli
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    i took the power rule so it is 2x and e^x is e^x

    • one year ago
  11. Chlorophyll
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    Then where is your PRODUCT rule for the first term?

    • one year ago
  12. ksaimouli
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    hmm ok |dw:1350749608453:dw|

    • one year ago
  13. ksaimouli
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    |dw:1350749664511:dw|

    • one year ago
  14. ksaimouli
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    is this correct

    • one year ago
  15. Chlorophyll
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    I'm lost, totally :( Now, you should follow my guiding: For the first term: u = x² --> u' = v = e^x --> v' = => u' v + uv' = ....

    • one year ago
  16. ksaimouli
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    |dw:1350750048815:dw|

    • one year ago
  17. ksaimouli
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    |dw:1350750099466:dw|

    • one year ago
  18. Chlorophyll
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    Yes, at least some light shining out of the tunnel here :)

    • one year ago
  19. ksaimouli
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    so whats next

    • one year ago
  20. ksaimouli
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    |dw:1350750252467:dw|

    • one year ago
  21. ksaimouli
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    so this is answer

    • one year ago
  22. Chlorophyll
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    Make it neat, so you can easily simplify them: ( x²e^x + 2x e^x ) - ( x e^x + e^x) = e^x ( x² + 2x - x - 1) = e^x ( x² + x - 1)

    • one year ago
  23. ksaimouli
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    i used the same formula for this one but i did not get

    • one year ago
  24. ksaimouli
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    |dw:1350750484771:dw|

    • one year ago
  25. Chlorophyll
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    Does it look simple enough?

    • one year ago
  26. ksaimouli
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    yup

    • one year ago
  27. ksaimouli
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    |dw:1350750529693:dw|

    • one year ago
  28. ksaimouli
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    this is wrong but i dont know where i did the mistake

    • one year ago
  29. Chlorophyll
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    So don't make it complicated by jumping all over, make each line is worthy effort :)

    • one year ago
  30. bahrom7893
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    x^2(e^x) - x(e^x) ... make your life simple, pull out e^x

    • one year ago
  31. bahrom7893
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    e^x * (x^2 - x), now use the product rule

    • one year ago
  32. bahrom7893
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    e^x * (2x-1) + (x^2 - x) * e^x

    • one year ago
  33. bahrom7893
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    Now simplify, pull out e^x again: e^x (2x - 1 + x^2 - x) = e^x (x^2 + x - 1)

    • one year ago
  34. bahrom7893
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    Now, you can either leave it like that, or distribute e^x. If you choose to do the latter, e^x * x^2 + e^x * x - e^x

    • one year ago
  35. ksaimouli
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    what about this one |dw:1350751090719:dw|

    • one year ago
  36. ksaimouli
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    @Chlorophyll

    • one year ago
  37. ksaimouli
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    if i take the product rule the e^2

    • one year ago
  38. ksaimouli
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    |dw:1350751424118:dw|

    • one year ago
  39. ksaimouli
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    can u plz help with this

    • one year ago
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