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jcamargo93 Group Title

Find dy/dx. y=5^(ln(x^2)+3)

  • one year ago
  • one year ago

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  1. abb0t Group Title
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    \[a^x \ln(a)\]

    • one year ago
  2. jcamargo93 Group Title
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    thank you!:)

    • one year ago
  3. abb0t Group Title
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    fa sho! GL!

    • one year ago
  4. RadEn Group Title
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    lol, the clue incomplete... use chain rule : y=x^2 ---> y'=2x v=ln(u) ---> v'=1/u w=5^v ---> w'=5^v * ln(5)

    • one year ago
  5. jcamargo93 Group Title
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    ok, i got this.. y'= 5^(ln((x^2)+3)) *ln 5*((2x)/((x^2)+3)) ..is it right?

    • one year ago
  6. RadEn Group Title
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    for the power of 5, is {ln(x^2) + 3} or ln(x^2+3)

    • one year ago
  7. jcamargo93 Group Title
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    i did include that, im just not sure if i did it right?, do i have to put (1/5) for ln5?

    • one year ago
  8. RadEn Group Title
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    before verify, i want to asking u is ur question |dw:1355302758870:dw|

    • one year ago
  9. RadEn Group Title
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    which one, the 1st or the 2nd ?

    • one year ago
  10. jcamargo93 Group Title
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    the 2nd one

    • one year ago
  11. RadEn Group Title
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    hohohoo... ok,, u have typo above, right ?

    • one year ago
  12. jcamargo93 Group Title
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    i think so,..

    • one year ago
  13. jcamargo93 Group Title
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    but yea, my question is to derive the 2nd one

    • one year ago
  14. RadEn Group Title
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    if u have typed ur question like the original above, the result will be difference :) OK,,,, now U RE RIGHT :*

    • one year ago
  15. jcamargo93 Group Title
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    so im right?

    • one year ago
  16. RadEn Group Title
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    sure... nice answer.

    • one year ago
  17. RadEn Group Title
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    but if ur question y=5^(ln(x^2)+3), so the answer will be y' = (2/x) * 5^(ln(x^2)+3) * ln(5)

    • one year ago
  18. jcamargo93 Group Title
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    ok then, well its like the second one

    • one year ago
  19. RadEn Group Title
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    yea... be careful in type the equation ^^

    • one year ago
  20. jcamargo93 Group Title
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    ok thanks!:)

    • one year ago
  21. RadEn Group Title
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    very welcome

    • one year ago
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