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Math Help - Derivative of ln(1+cosx)

  1. #1
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    Derivative of ln(1+cosx)

    Dear All!
    Please, if the f'(1+cosx)=1/(1+cosx) or sinx/(1+cosx)?
    I would say the first one, but got the second one in a book. Is that a typo?
    Many thanks!
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    Re: Derivative of ln(1+cosx)

    \frac{d}{dx} \left[\ln(1+\cos{x})\right] = -\frac{\sin{x}}{1+\cos{x}}
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    Re: Derivative of ln(1+cosx)

    What kind of courses are you taking or have taken? If you are asked about derivatives, you should be taking Calculus- but then you should surely know that "f'(ln(1+ cos(x))" is just bad notation. You mean "if f(x)= ln(1+ cos(x)), what is f'(x)?" Now, do you know the chain rule? Clearly, you are expected to because it is needed here. You have f(u)= ln(u) with u= 1+ cos(x). Then \frac{df}{dx}= \frac{df}{du}\frac{du}{dx}. What is the derivative of ln(u) with respect to u? What is the derivative of 1+ cos(x) with respect to x?
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    Re: Derivative of ln(1+cosx)

    Thank you!
    But what is procedure to get it?
    Many thanks
    Last edited by marijakopljar; December 29th 2012 at 08:14 AM.
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    Re: Derivative of ln(1+cosx)

    Remember, always differentiate from the outside first.
    When you differentiate lnx, you'll end up with \frac{1}{x}
    So, when you differentiate ln(1+\cos\theta), you will get \frac{1}{1+\cos\theta}
    Next, differentiate within the brackets, ie. (1+\cos\theta)
    Differentiate 1, it will be 0. Differentiate \cos\theta you'll end up with \-sin\theta
    You have -\sin\theta and \frac{1}{1+\cos\theta},
    multiplying,
    \frac{-\sin\theta}{1+\cos\theta}
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    Re: Derivative of ln(1+cosx)

    u is a function of x ...

    \frac{d}{dx} [\ln(u)] = \frac{u'}{u}
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    Re: Derivative of ln(1+cosx)

    Tutu and skeeter are both referring to the chain rule. That is a very important rule for differentiation- look it up in your Calculus text.
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    Re: Derivative of ln(1+cosx)

    Thank you very muc-it become clear in just a few minutes!
    Please, what if we have derivative of ln(sinx+cosx)?

    I supose, we first have 1/(sinx+cosx)
    then we have cosx and (-sinx)
    so we multiply 1/(sinx+cosx) with (cosx-sinx) and we get (cosx-sinx)/(sinx+cosx)
    Am I right?
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    Re: Derivative of ln(1+cosx)

    Quote Originally Posted by marijakopljar View Post

    I supose, we first have 1/(sinx+cosx)
    then we have cosx and (-sinx)
    so we multiply 1/(sinx+cosx) with (cosx-sinx) and we get (cosx-sinx)/(sinx+cosx)
    Am I right?
    YES!
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  10. #10
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    Re: Derivative of ln(1+cosx)

    Quote Originally Posted by marijakopljar View Post
    Please,
    if we have:
    lim(x to 1) ((x/(x-1)-1/lnx)
    then we get:

    lim(x to 1) ((xlnx-x+1)/(xlnx-lnx))
    then we take x=1:

    lim(x to1)((ln(1)+1-1) / (ln(1)-ln(1))
    we get 0/0
    and can apply LH:

    but NOW:
    How do we get NOW lim(x to1)((xlnx)/(xlnx+x-1)???
    What is the inversion doing here?
    Many thanks!

    in future, start a new problem with a new thread ... I have moved this post and re-titled it as Limit Problem
    Last edited by skeeter; December 30th 2012 at 05:33 AM.
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