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monroe17

  • 3 years ago

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

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

  2. dpaInc
    • 3 years ago
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    yuck!!!

  3. monroe17
    • 3 years ago
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    I know ;(

  4. dpaInc
    • 3 years ago
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    i think it would be better to do logarithmic differentiation...

  5. dpaInc
    • 3 years ago
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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...

  6. dpaInc
    • 3 years ago
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    i guess that's not so hard afterall....

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

  8. dpaInc
    • 3 years ago
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    use the logaritm properties...

  9. monroe17
    • 3 years ago
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    I need to look those up ;/ and review them.

  10. dpaInc
    • 3 years ago
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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...

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

  12. dpaInc
    • 3 years ago
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    yeah... should be easier now....

  13. monroe17
    • 3 years ago
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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}

  14. monroe17
    • 3 years ago
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    is it..ln(x^4)=4lnx

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

  16. dpaInc
    • 3 years ago
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    that's the first step...

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

  18. dpaInc
    • 3 years ago
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    ooo... sorry my bad... the second part is wrong....

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

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

  21. dpaInc
    • 3 years ago
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    and yes (to your last post)

  22. dpaInc
    • 3 years ago
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    lemme rewrite what i shouldve written....

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

  24. monroe17
    • 3 years ago
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    \[\frac{ 4 }{ x }+3..\]

  25. dpaInc
    • 3 years ago
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    yep... keep going...

  26. monroe17
    • 3 years ago
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    I need to do chain rule?

  27. dpaInc
    • 3 years ago
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    yes, chain rule: \(\large [lny]'=\frac{y'}{y} \)

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

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

  30. dpaInc
    • 3 years ago
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    haha... looks good..:)

  31. monroe17
    • 3 years ago
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    why haha? did I do something wrong?

  32. dpaInc
    • 3 years ago
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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...

  33. monroe17
    • 3 years ago
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    oh lol :) you just discouraged me! >;[ no im jk ;) thank you!

  34. dpaInc
    • 3 years ago
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    oh wait...

  35. dpaInc
    • 3 years ago
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    haha... just kidding.... yw... :)

  36. monroe17
    • 3 years ago
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    haha! jerk ;D

  37. dpaInc
    • 3 years ago
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    :)

  38. dpaInc
    • 3 years ago
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    i forgot... nice work on your part... way to stick with the problem....

  39. monroe17
    • 3 years ago
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    what you mean? I always finish ;D

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