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How to find the antiderivative of cot(x)

  1. Jan 11, 2009 #1
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    5) As you can see ln(sin(1)) +1/ln.5 does not yield one of the above answers. Can someone tell me where I went wrong?
     
  2. jcsd
  3. Jan 11, 2009 #2
    Re: antiderivative(cotx)dx

    I think you made a mistake when finding C.
    C should be equal to 1-ln(.5) not 1/ln(.5).
    The answer should be the last one (E).
     
  4. Jan 11, 2009 #3
    Re: antiderivative(cotx)dx

    Thanks for the help, didn't see that.
     
  5. Jan 11, 2009 #4
    Re: antiderivative(cotx)dx

    Does anyone have an idea as to how to figure 6 out? I don't believe I have ever tried one of those problems.
     
  6. Jan 11, 2009 #5
    Re: antiderivative(cotx)dx

    You should try finding out what is the second derivative of [f^2(x)].
    When you're done, you just plug in the number 3 and it is easy to find out what's the answer.
     
  7. Jan 11, 2009 #6
    Re: antiderivative(cotx)dx

    isn't the second derivative of f^2x = 2?
     
  8. Jan 11, 2009 #7
    Re: antiderivative(cotx)dx

    No, consider [f^2(x)]=[f(x)]^2
    you know that (u^m)'=m (u^(m-1)) u' right?
    So it's the same thing for u=f(x) and m=2.

    Can you try it and see what you get when you do it twice?
     
  9. Jan 11, 2009 #8
    Re: antiderivative(cotx)dx

    Certainly not. Let [tex]f(x) = x^3[/tex], then [tex]f^{2}(x) = x^6[/tex]. Still think your rule works?
     
  10. Jan 11, 2009 #9
    Re: antiderivative(cotx)dx

    I can't see how to take the second derivative of f^2(x). Am I supposed to use the previous f(3) = 2 and ect for this? I have never actually learned how.
     
  11. Jan 11, 2009 #10
    Re: antiderivative(cotx)dx

    Use the chain rule, if I told you to differentiate [sin(x)]^2 what would you do?
     
  12. Jan 11, 2009 #11
    Re: antiderivative(cotx)dx

    so we start with

    f(x)^2
    2f(x)
    and then what?
     
  13. Jan 11, 2009 #12
    Re: antiderivative(cotx)dx

    No, that's not it. Try again :) You are saying if we define f(x) = sin(x) then f^(2)(x) = sin^2(x) and you are saying derivative of f^(2)(x) is 2*sin(x)? You know that's not true...
     
  14. Jan 11, 2009 #13
    Re: antiderivative(cotx)dx

    could I possibly multiply them and get f(2x^2)?
     
  15. Jan 11, 2009 #14
    Re: antiderivative(cotx)dx

    [f(x)^2]'= 2f(x)*f '(x)
    you're not using the chain rule correctly. Don't forget the f '(x) at the end :)

    Now can you find out what is [2f(x)*f '(x)] ' using the product rule?
     
  16. Jan 11, 2009 #15
    Re: antiderivative(cotx)dx

    Where are you getting 2x^2 from? Think of f^2(x) as f(x)*f(x). What's the derivative of that. That uses the product rule instead of the chain rule, if you have trouble remembering that one.
     
  17. Jan 11, 2009 #16
    Re: antiderivative(cotx)dx

    2f'(x) + f"(x)2f(x) would that be it and then just number plug?
     
  18. Jan 11, 2009 #17
    Re: antiderivative(cotx)dx

    No! Why is the first term 2f'(x)?
     
  19. Jan 11, 2009 #18
    Re: antiderivative(cotx)dx

    it should be 2f'(x)f'(x) + f"(x)2f(x) = 42. I messed up the chain rule again :(
     
    Last edited: Jan 11, 2009
  20. Jan 11, 2009 #19
    Re: antiderivative(cotx)dx

    Now 7 is another problem I have yet to face. I see that t is in seconds so I think I should multiply t by 60 to convert it to minutes. Should I then plug 1 in for y and take the derivative?
     
  21. Jan 11, 2009 #20
    Re: antiderivative(cotx)dx

    Actually that's the product rule.

    If [tex] \frac{dy}{dt} = ky [/tex] Then y = ? Solve this separable DE.
     
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