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Please check my derivative

  1. Feb 28, 2013 #1
    I'm preparing for a test next week and am working on some practice problems we got, but we did not get the answers. There are 2 problems where I'm not sure if I have done it correct or not and would like to have it checked.

    This is the first problem. I think I have followed the chain, product and exponential rules, but I really have no idea if this is correct or not.
    attachment.php?attachmentid=56219&stc=1&d=1362116653.png

    Any help is appreciated.
     

    Attached Files:

  2. jcsd
  3. Feb 28, 2013 #2

    Mute

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    Yep, looks correct.
     
  4. Feb 28, 2013 #3
    wow, thank you very much for the quick reply
     
  5. Mar 1, 2013 #4
    Instead of starting a new thread, I will post the second problem here.
    I think I have done this one correct, but as with the first problem, I really have no idea if it is correct or not
    attachment.php?attachmentid=56236&stc=1&d=1362148368.png
     

    Attached Files:

  6. Mar 1, 2013 #5

    haruspex

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    That's correct, but you could go further and use the second line of that to substitute for y in the last line, so you have y' entirely in terms of x. Don't know if that's what was wanted.
     
  7. Mar 1, 2013 #6

    Mark44

    Staff: Mentor

    In the future, when you post a new problem, please start a new thread.
     
  8. Mar 1, 2013 #7
    >>In the future, when you post a new problem, please start a new thread
    ok, I will remember that

    >>That's correct, but you could go further and
    Good point. But in this case since I don't start out with a y = ... , but rather yx = ..., then I'm not sure if it would be expected.
    If I were to replace it, I should replace y for e[itex]\frac{tan(x) ln(x)}{x}[/itex], correct?.
    Anyway, I will try and keep this in mind for the test.
    Thank you very much for your help
     
  9. Mar 1, 2013 #8

    haruspex

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    Yes, or ##y = x^{\frac{tan(x)}{x}}##. You could reduce it to ##y' = x^{\frac{tan(x)}{x}-1}[\frac{tan(x)}{x}\left(1-ln(x)\right)+ln(x)sec^2(x)]##
     
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