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hooverst Group Title

y=e^xlnx y=ln(3x^2) y=lnx/x^2+1

  • 2 years ago
  • 2 years ago

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  1. dpaInc Group Title
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    what's the question? which equation is nicest?

    • 2 years ago
  2. dpflan Group Title
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    What's the goal? Solve for x in this system of equations?

    • 2 years ago
  3. dpflan Group Title
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    heh, I like the first one ;)

    • 2 years ago
  4. dpaInc Group Title
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    me too! :)

    • 2 years ago
  5. hooverst Group Title
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    each one is a different question its says compute the derivatives of each function

    • 2 years ago
  6. dpflan Group Title
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    Oh, OK, are you familiar with (1.) product rule and (2.) quotient rule and the (3.) power rule?

    • 2 years ago
  7. hooverst Group Title
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    yeah a little

    • 2 years ago
  8. dpflan Group Title
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    (1.) \[y=e^x*ln(x)\] (2.) \[y=ln(3x^2)\] (3.) \[y=\frac{ln(x)}{x^2}+1\]

    • 2 years ago
  9. dpflan Group Title
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    And you know chain rule? a little, right? ;)

    • 2 years ago
  10. hooverst Group Title
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    the 1st equation is \[y=e ^{xlnx}\]

    • 2 years ago
  11. hooverst Group Title
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    yes

    • 2 years ago
  12. dpflan Group Title
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    OK, well, let's work with equation 2, for me that is the easiest one. We are differentiating with respect to y, right?

    • 2 years ago
  13. hooverst Group Title
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    yes

    • 2 years ago
  14. dpflan Group Title
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    Cool, \[y=ln(3*x^2)\] We will use the Chain Rule here

    • 2 years ago
  15. dpflan Group Title
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    \[\frac{d}{dx}(ln(3*x^2)) = \frac{d*ln(u)}{du}*\frac{du}{dx}\] \[u = 3*x^2\]and \[\frac{d*ln(u)}{du} = \frac{1}{u}\] \[\frac{du}{dx} = 3*2*x^{2-1} = 6x\]

    • 2 years ago
  16. dpflan Group Title
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    Yes, and to get du/dx we use the power rule

    • 2 years ago
  17. dpflan Group Title
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    \[x^y = y*x^{y-1}\] is the power rule

    • 2 years ago
  18. dpflan Group Title
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    Can you finish solving the equation?

    • 2 years ago
  19. hooverst Group Title
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    ok cause when he explained it in class i did not understand it at all.

    • 2 years ago
  20. dpflan Group Title
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    All right, how is it now?

    • 2 years ago
  21. hooverst Group Title
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    ok so for the power rule what do i plug in or do I have to plug anything in

    • 2 years ago
  22. dpflan Group Title
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    Nah, you don't need to plug-in anything. So, what do you have so far? If you are writing it down, you can snap a picture with your phone if it has a camera, then post the picture here

    • 2 years ago
  23. dpflan Group Title
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    \[\frac{d}{dx}(ln(3*x^2)) = \frac{d*ln(u)}{du}*\frac{du}{dx}\] \[\frac{d*ln(u)}{du} = \frac{1}{u}\] \[\frac{du}{dx} = 3*2*x^{2-1} = 6x\] \[u=3*x^2\] So you have \[\frac{1}{3*x^2} * 6x\] Simplify and you're done with that equation

    • 2 years ago
  24. hooverst Group Title
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    O ok I didn't know whether what I had was right but thats what I just got to.

    • 2 years ago
  25. dpflan Group Title
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    This maybe the hardest of the bunch because of the substitution, chain rule, and product rule

    • 2 years ago
  26. hooverst Group Title
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    yea It was

    • 2 years ago
  27. dpflan Group Title
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    Well, because substitution can be confusing because you introduce a new variable to differentiate on

    • 2 years ago
  28. dpflan Group Title
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    *or with respect to

    • 2 years ago
  29. hooverst Group Title
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    O ok

    • 2 years ago
  30. dpflan Group Title
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    Now (1.)\[y=e^x * lnx\]

    • 2 years ago
  31. Callisto Group Title
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    Excuse ... me.... are we finding dy/dx or dx/dy?

    • 2 years ago
  32. dpflan Group Title
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    dy/dx I believe

    • 2 years ago
  33. Callisto Group Title
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    Thank you. Please continue....

    • 2 years ago
  34. dpflan Group Title
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    Heh, thanks, are you following along @Callisto ?

    • 2 years ago
  35. dpflan Group Title
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    \[y=e^x* lnx\] Here we use the product rule. We can think of this equation as the product of 2 functions of x

    • 2 years ago
  36. dpflan Group Title
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    \[(f*g)' = f'*g + f*g'\] So the derivative of the product of these functions, f and g Is the sum of the derivative of the first * the second + the first * the derivative of the second

    • 2 years ago
  37. Callisto Group Title
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    For the second one, I was thinking in this way.. \[y=ln(3x^2)=ln3+ lnx^2 = ln3+2lnx\]\[\frac{dy}{dx}=\frac{d}{dx}2lnx = ...\]

    • 2 years ago
  38. dpflan Group Title
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    Yes, that is easier! @hooverst If you did it that way you don't have do substitution

    • 2 years ago
  39. hooverst Group Title
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    Ok dpflan

    • 2 years ago
  40. dpflan Group Title
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    It can be really fun in math the find the simpler way of doing things, so you just use some intuition and understanding to make your work easier, simpler can be much more beautiful too

    • 2 years ago
  41. dpflan Group Title
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    *to find

    • 2 years ago
  42. dpflan Group Title
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    Nice one @Callisto @hooverst So, you think you can solve equation (1.) with the product rule?

    • 2 years ago
  43. dpaInc Group Title
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    hold on... i think @hooverst clarified that the first equation is: \(\large y=e^{xlnx} \) , not \(\large y=e^{x}\cdot lnx \)

    • 2 years ago
  44. dpflan Group Title
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    Heh, you're right, man

    • 2 years ago
  45. hooverst Group Title
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    so callisto when you solved your equation for the second 1 what did you get for your anserw

    • 2 years ago
  46. dpflan Group Title
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    Thanks, getting a little carried away. Let me step away for a bit

    • 2 years ago
  47. dpaInc Group Title
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    ur doing fine....:)

    • 2 years ago
  48. Callisto Group Title
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    @hooverst What did you get ??

    • 2 years ago
  49. hooverst Group Title
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    is the anserw \[y=\ln(3x^2)=1\div 3x^2\]

    • 2 years ago
  50. Callisto Group Title
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    Not really... what is \(\frac{d}{dx}lnx\) ?

    • 2 years ago
  51. hooverst Group Title
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    im not sure

    • 2 years ago
  52. dpaInc Group Title
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    are we still working on the second equation? @hooverst ???

    • 2 years ago
  53. dpaInc Group Title
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    because you can do this many different ways... but in the end, the derivative should be the same...

    • 2 years ago
  54. hooverst Group Title
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    yes but I was trying to figure out was 2lnx the anserw for #2 or is y=ln(3x^2)=1/3x^2 the anserw

    • 2 years ago
  55. dpaInc Group Title
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    answer as in the derivative? you and @dpflan , worked it out to \(\large \frac{6x}{3x^2} \) simplified to....???

    • 2 years ago
  56. dpaInc Group Title
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    in the case with @Callisto , her method was to simplify the function first: \(\large y=ln(3x^2)=ln3+2lnx \) so \(\large y'=[ln3]'+[2lnx]'= \) ...???

    • 2 years ago
  57. dpaInc Group Title
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    and either way the derivative is the same...

    • 2 years ago
  58. Callisto Group Title
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    The key point is you need to know what \(\frac{d}{dx}lnx\) is.

    • 2 years ago
  59. dpflan Group Title
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    Yes, that is a derivative you need to memorize, it will be quite useful

    • 2 years ago
  60. hooverst Group Title
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    Ok so many of you have said had your own opinion about the equation so which one is the right 1.

    • 2 years ago
  61. dpflan Group Title
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    Hehe, right, so, what is your opinion? ;p You can solve a problem many, many, different ways

    • 2 years ago
  62. dpflan Group Title
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    For you, how would you approach it now that you've seen how we would?

    • 2 years ago
  63. Callisto Group Title
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    I would say, none of us have given you the final answer. But we have given you some steps you need using different approaches. Though, the final answer will be the same.

    • 2 years ago
  64. hooverst Group Title
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    Ok Think the one Callisto gave me is more simple for #2

    • 2 years ago
  65. dpflan Group Title
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    Definitely, that was an awesome application of intuition

    • 2 years ago
  66. dpflan Group Title
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    well, and mathematical understanding

    • 2 years ago
  67. dpaInc Group Title
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    i guess it comes down to preference because i personally would've used u'/u ... the method used earlier...

    • 2 years ago
  68. Callisto Group Title
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    The step I left for you is to find \(\frac{d}{dx}lnx\). Multiply the answer you get for that by 2. Then, it's done.

    • 2 years ago
  69. hooverst Group Title
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    ok

    • 2 years ago
  70. Callisto Group Title
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    May I ask you again- what is d/dx ( lnx) ?

    • 2 years ago
  71. hooverst Group Title
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    would you multiply it.

    • 2 years ago
  72. Callisto Group Title
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    Nope.... Hint: look it up in the pdf.: http://tutorial.math.lamar.edu/pdf/Calculus_Cheat_Sheet_Derivatives.pdf You can find the answer for d/dx (lnx) there.

    • 2 years ago
  73. hooverst Group Title
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    Ok is it d/dx(ln(x))=1/x

    • 2 years ago
  74. Callisto Group Title
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    Yes. dy/dx = d/dx (ln3 + lnx^2) = d/dx (2lnx) = 2 d/dx(lnx) = ...?

    • 2 years ago
  75. hooverst Group Title
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    is it 2/x or just 2x Idon't know if its right

    • 2 years ago
  76. Callisto Group Title
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    Which one do you think? (i) 2/x (ii) 2x

    • 2 years ago
  77. hooverst Group Title
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    2/x

    • 2 years ago
  78. Callisto Group Title
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    Yes. That's correct. Any questions?

    • 2 years ago
  79. hooverst Group Title
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    just 1 when we started simplifying the equation where did you get the 2 from?

    • 2 years ago
  80. Callisto Group Title
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    \[lnx^a = a\ln x\]

    • 2 years ago
  81. hooverst Group Title
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    Ok I get it. thanks Callisto

    • 2 years ago
  82. Callisto Group Title
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    Welcome.

    • 2 years ago
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