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hooverst

  • 2 years ago

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

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  1. dpaInc
    • 2 years ago
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    what's the question? which equation is nicest?

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

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

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

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

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

  7. hooverst
    • 2 years ago
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    yeah a little

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

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

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

  11. hooverst
    • 2 years ago
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    yes

  12. dpflan
    • 2 years ago
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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?

  13. hooverst
    • 2 years ago
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    yes

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

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

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

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

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

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

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

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

  22. dpflan
    • 2 years ago
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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

  23. dpflan
    • 2 years ago
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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

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

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

  26. hooverst
    • 2 years ago
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    yea It was

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

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

  29. hooverst
    • 2 years ago
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    O ok

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

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

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

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

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

  35. dpflan
    • 2 years ago
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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

  36. dpflan
    • 2 years ago
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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

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

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

  39. hooverst
    • 2 years ago
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    Ok dpflan

  40. dpflan
    • 2 years ago
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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

  41. dpflan
    • 2 years ago
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    *to find

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

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

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

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

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

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

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

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

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

  51. hooverst
    • 2 years ago
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    im not sure

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

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

  54. hooverst
    • 2 years ago
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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

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

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

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

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

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

  60. hooverst
    • 2 years ago
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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.

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

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

  63. Callisto
    • 2 years ago
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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.

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

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

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

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

  68. Callisto
    • 2 years ago
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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.

  69. hooverst
    • 2 years ago
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    ok

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

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

  72. Callisto
    • 2 years ago
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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.

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

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

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

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

  77. hooverst
    • 2 years ago
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    2/x

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

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

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

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

  82. Callisto
    • 2 years ago
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    Welcome.

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