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The derivative of ln cos x

  1. Feb 26, 2012 #1
    1. The problem statement, all variables and given/known data

    the question is for plane curves. find T N and k for the plane curves

    1. r(t) = ti + ln (cos t)

    the derivative of ln x is 1/|x|, so why isn't the derivative of ln cos x, 1/cos x or sec x?

    My book says the answer is - tan x
     
    Last edited: Feb 26, 2012
  2. jcsd
  3. Feb 26, 2012 #2
    Because [itex]\ln{\cos{x}}[/itex] is actually a composition of the functions [itex]\ln{x}[/itex] and [itex]\cos{x}[/itex]. Therefore you need to use the chain rule.
     
  4. Feb 26, 2012 #3
    I put in the question into the original post. I don't see how they're using the chain rule. They aren't deriving two terms, it's just ln (cos t). if they're driving t, then i would think the derivative of t would just be 1.
     
  5. Feb 26, 2012 #4
    you need to use the derivative of ln r(t), which is r'(t)/ r (t)
    try that!
     
  6. Feb 26, 2012 #5

    vela

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    Consider this problem: Find the derivative of ##y=\ln t^2##.

    First method: Using the property of logs, we pull the exponent out front.
    $$y=\ln t^2 = 2\ln t$$ When you differentiate this, you get
    $$y' = 2\left(\frac{1}{t}\right).$$ This is the correct answer.

    Second method: Just "differentiate" the way you did
    $$y' = \frac{1}{t^2}.$$ This is wrong. Why is it wrong?
     
  7. Feb 26, 2012 #6
    but there's no exponent involved here. plus look at this problem: they just derive everything normally here. they don't do the r'/r

    [​IMG]
     
  8. Feb 26, 2012 #7

    I like Serena

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    The chain rule says that the derivative of f(g(t)) is f'(g(t))g'(t).
    This is what vela illustrated.

    In your case, for ln(cos(t)), you have calculated f'(g(t)) which is 1/cos(t), but you still need to multiply by g'(t) which is the derivative of cos(t).
     
  9. Feb 26, 2012 #8
    ok, I sort of understand.
     
  10. Feb 26, 2012 #9

    I like Serena

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    So do you understand how the book got -tan(t)?
     
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