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hooverst

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

  • one year ago
  • one year ago

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

    • one year ago
  2. dpflan
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    What's the goal? Solve for x in this system of equations?

    • one year ago
  3. dpflan
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    heh, I like the first one ;)

    • one year ago
  4. dpaInc
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    me too! :)

    • one year ago
  5. hooverst
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    each one is a different question its says compute the derivatives of each function

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

    • one year ago
  7. hooverst
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    yeah a little

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

    • one year ago
  9. dpflan
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    And you know chain rule? a little, right? ;)

    • one year ago
  10. hooverst
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    the 1st equation is \[y=e ^{xlnx}\]

    • one year ago
  11. hooverst
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    yes

    • one year ago
  12. dpflan
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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?

    • one year ago
  13. hooverst
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    yes

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

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

    • one year ago
  16. dpflan
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    Yes, and to get du/dx we use the power rule

    • one year ago
  17. dpflan
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    \[x^y = y*x^{y-1}\] is the power rule

    • one year ago
  18. dpflan
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    Can you finish solving the equation?

    • one year ago
  19. hooverst
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    ok cause when he explained it in class i did not understand it at all.

    • one year ago
  20. dpflan
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    All right, how is it now?

    • one year ago
  21. hooverst
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    ok so for the power rule what do i plug in or do I have to plug anything in

    • one year ago
  22. dpflan
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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

    • one year ago
  23. dpflan
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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

    • one year ago
  24. hooverst
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    O ok I didn't know whether what I had was right but thats what I just got to.

    • one year ago
  25. dpflan
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    This maybe the hardest of the bunch because of the substitution, chain rule, and product rule

    • one year ago
  26. hooverst
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    yea It was

    • one year ago
  27. dpflan
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    Well, because substitution can be confusing because you introduce a new variable to differentiate on

    • one year ago
  28. dpflan
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    *or with respect to

    • one year ago
  29. hooverst
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    O ok

    • one year ago
  30. dpflan
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    Now (1.)\[y=e^x * lnx\]

    • one year ago
  31. Callisto
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    Excuse ... me.... are we finding dy/dx or dx/dy?

    • one year ago
  32. dpflan
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    dy/dx I believe

    • one year ago
  33. Callisto
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    Thank you. Please continue....

    • one year ago
  34. dpflan
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    Heh, thanks, are you following along @Callisto ?

    • one year ago
  35. dpflan
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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

    • one year ago
  36. dpflan
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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

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

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

    • one year ago
  39. hooverst
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    Ok dpflan

    • one year ago
  40. dpflan
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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

    • one year ago
  41. dpflan
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    *to find

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

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

    • one year ago
  44. dpflan
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    Heh, you're right, man

    • one year ago
  45. hooverst
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    so callisto when you solved your equation for the second 1 what did you get for your anserw

    • one year ago
  46. dpflan
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    Thanks, getting a little carried away. Let me step away for a bit

    • one year ago
  47. dpaInc
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    ur doing fine....:)

    • one year ago
  48. Callisto
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    @hooverst What did you get ??

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

    • one year ago
  50. Callisto
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    Not really... what is \(\frac{d}{dx}lnx\) ?

    • one year ago
  51. hooverst
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    im not sure

    • one year ago
  52. dpaInc
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    are we still working on the second equation? @hooverst ???

    • one year ago
  53. dpaInc
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    because you can do this many different ways... but in the end, the derivative should be the same...

    • one year ago
  54. hooverst
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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

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

    • one year ago
  56. dpaInc
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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 \(\large y'=[ln3]'+[2lnx]'= \) ...???

    • one year ago
  57. dpaInc
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    and either way the derivative is the same...

    • one year ago
  58. Callisto
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    The key point is you need to know what \(\frac{d}{dx}lnx\) is.

    • one year ago
  59. dpflan
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    Yes, that is a derivative you need to memorize, it will be quite useful

    • one year ago
  60. hooverst
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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.

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

    • one year ago
  62. dpflan
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    For you, how would you approach it now that you've seen how we would?

    • one year ago
  63. Callisto
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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.

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

    • one year ago
  65. dpflan
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    Definitely, that was an awesome application of intuition

    • one year ago
  66. dpflan
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    well, and mathematical understanding

    • one year ago
  67. dpaInc
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    i guess it comes down to preference because i personally would've used u'/u ... the method used earlier...

    • one year ago
  68. Callisto
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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.

    • one year ago
  69. hooverst
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    ok

    • one year ago
  70. Callisto
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    May I ask you again- what is d/dx ( lnx) ?

    • one year ago
  71. hooverst
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    would you multiply it.

    • one year ago
  72. Callisto
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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.

    • one year ago
  73. hooverst
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    Ok is it d/dx(ln(x))=1/x

    • one year ago
  74. Callisto
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    Yes. dy/dx = d/dx (ln3 + lnx^2) = d/dx (2lnx) = 2 d/dx(lnx) = ...?

    • one year ago
  75. hooverst
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    is it 2/x or just 2x Idon't know if its right

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

    • one year ago
  77. hooverst
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    2/x

    • one year ago
  78. Callisto
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    Yes. That's correct. Any questions?

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

    • one year ago
  80. Callisto
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    \[lnx^a = a\ln x\]

    • one year ago
  81. hooverst
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    Ok I get it. thanks Callisto

    • one year ago
  82. Callisto
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    Welcome.

    • one year ago
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