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monroe17 Group Title

Find dy/dx step by step? please? :)

  • one year ago
  • one year ago

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  1. monroe17 Group Title
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    \[y=ln(\frac{ x^4e^{3x} }{ \sqrt{2x^5-1} })\]

    • one year ago
  2. dpaInc Group Title
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    yuck!!!

    • one year ago
  3. monroe17 Group Title
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    I know ;(

    • one year ago
  4. dpaInc Group Title
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    i think it would be better to do logarithmic differentiation...

    • one year ago
  5. dpaInc Group Title
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    \(\large y=ln(\frac{x^4e^{3x}}{(2x-5)^{\frac{1}{2}}}) = 4lnx + 3x - \frac{1}{2}(2x^5-1)\) simplifying y that way should be much simpler...

    • one year ago
  6. dpaInc Group Title
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    i guess that's not so hard afterall....

    • one year ago
  7. monroe17 Group Title
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    wait wait.. how'd you get 4ln(x)+3x-1/2(2x^5-1)

    • one year ago
  8. dpaInc Group Title
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    use the logaritm properties...

    • one year ago
  9. monroe17 Group Title
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    I need to look those up ;/ and review them.

    • one year ago
  10. dpaInc Group Title
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    \(\large ln(AB)=lnA+lnB \); \(\large ln(A/B)=lnA-lnB \); \(\large ln(A^B)=BlnA \); hmm... i think that's all of 'em...

    • one year ago
  11. monroe17 Group Title
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    Awe, well thank you. Okay so now back to what you got.. Let me look over that real quick

    • one year ago
  12. dpaInc Group Title
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    yeah... should be easier now....

    • one year ago
  13. monroe17 Group Title
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    Okay so I see you used the second one.. But, I'm confused how you got those values... ln(A)=4lnx+3x when A= x^4e^{3x}

    • one year ago
  14. monroe17 Group Title
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    is it..ln(x^4)=4lnx

    • one year ago
  15. dpaInc Group Title
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    \(\large y=ln(\frac{x^4e^{3x}}{(2x-5)^{\frac{1}{2}}}) =ln(x^4e^{3x})-ln(2x-5)^{1/2} \)

    • one year ago
  16. dpaInc Group Title
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    that's the first step...

    • one year ago
  17. dpaInc Group Title
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    now.... \(\large ln(x^4e^{3x})=lnx^4+lne^{3x}=4lnx+3x \) do u undestand this part?

    • one year ago
  18. dpaInc Group Title
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    ooo... sorry my bad... the second part is wrong....

    • one year ago
  19. monroe17 Group Title
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    ohhh... ln(A^b)=b*ln(A) ln(x^4)=4*ln(x) and e&ln cancel out leaving 3x

    • one year ago
  20. dpaInc Group Title
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    \(\large ln(2x-5)^{1/2}=\frac{1}{2}ln(2x-5) \) that's what it should have been...

    • one year ago
  21. dpaInc Group Title
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    and yes (to your last post)

    • one year ago
  22. dpaInc Group Title
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    lemme rewrite what i shouldve written....

    • one year ago
  23. dpaInc Group Title
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    \(\large y=ln(\frac{x^4e^{3x}}{(2x-5)^{\frac{1}{2}}}) = 4lnx + 3x - \frac{1}{2}\color {red}{ln}(2x^5-1) \)

    • one year ago
  24. monroe17 Group Title
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    \[\frac{ 4 }{ x }+3..\]

    • one year ago
  25. dpaInc Group Title
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    yep... keep going...

    • one year ago
  26. monroe17 Group Title
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    I need to do chain rule?

    • one year ago
  27. dpaInc Group Title
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    yes, chain rule: \(\large [lny]'=\frac{y'}{y} \)

    • one year ago
  28. dpaInc Group Title
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    so \(\large [ln(2x^5-1)]'=\frac{[2x^5-1]'}{2x^5-1} \) simplify....

    • one year ago
  29. monroe17 Group Title
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    \[\frac{ 4 }{ x }+3-\frac{ 1 }{ 2 }\frac{ 10x^4 }{ 2x^5-1 } = \frac{ 4 }{ x }+3-\frac{ 5x^4 }{ 2x^5-1 }\]

    • one year ago
  30. dpaInc Group Title
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    haha... looks good..:)

    • one year ago
  31. monroe17 Group Title
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    why haha? did I do something wrong?

    • one year ago
  32. dpaInc Group Title
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    no... i was getting the wrong answers until at the last moment when u posted ur answer... it confirmed my last answer...

    • one year ago
  33. monroe17 Group Title
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    oh lol :) you just discouraged me! >;[ no im jk ;) thank you!

    • one year ago
  34. dpaInc Group Title
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    oh wait...

    • one year ago
  35. dpaInc Group Title
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    haha... just kidding.... yw... :)

    • one year ago
  36. monroe17 Group Title
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    haha! jerk ;D

    • one year ago
  37. dpaInc Group Title
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    :)

    • one year ago
  38. dpaInc Group Title
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    i forgot... nice work on your part... way to stick with the problem....

    • one year ago
  39. monroe17 Group Title
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    what you mean? I always finish ;D

    • one year ago
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