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Math Help - Derivate.

  1. #1
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    Derivate.

    given is y(x)=ln(tan(\frac{x}{2}))-atan(x)ln(1+sin(x))-x

    so I would like to derivate this

    thus we can split it in 3 part

    First ln(tan(\frac{x}{2})) wil be \frac{1}{2}\frac{sec^2(\frac{x}{2})}{tan(\frac{x}{  2})}

    the second and third one atan(x)ln(1+sin(x))-x will be \frac{ln(1+sin(x))}{sin^2(x)}-\frac{atan(x)cos(x)}{1+sin(x)}-1

    finelay we get \frac{1}{2}\frac{sec^2(\frac{x}{2})}{tan(\frac{x}{  2})} - \frac{ln(1+sin(x))}{sin^2(x)}-\frac{atan(x)cos(x)}{1+sin(x)}-1

    Who can I simplify this? Greets Thanks a lot.
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  2. #2
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    Hello, Bert!

    An error in your derivative (probably a typo)
    . . Otherwise, excellent work!


    Differentiate: . y(x)\;=\;\ln(\tan\frac{x}{2}) - \arctan(x)\cdot\ln(1 + \sin x ) - x

    You should get: . \frac{1}{2}\cdot\frac{\sec^2\frac{x}{2}}{\tan\frac  {x}{2}} - \frac{\ln(1+\sin x)}{\underbrace{1 + x^2}}-\frac{\arctan(x)\cdot\cos x}{1 + \sin x }-1

    With three (four!) different denominators, there's not much chance of simplifying.


    The first term can be simplified a bit:

    . . \frac{1}{2}\cdot\frac{\frac{1}{\cos^2\frac{x}{2}}}  {\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}}\;=\; \frac{1}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \;\frac{1}{\sin x} \;=\;\csc x<br />
. . . but that's all . . .

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  3. #3
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    oké but the answer in my book say f'(x)=\frac{ln(1+sin(x))}{sin^2(x)}

    What is the problem ? Greets Thanks .
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  4. #4
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    Hello again, Bert!

    but the answer in my book say f'(x)=\frac{\ln(1+\sin x )}{\sin^2 x }

    What is the problem ?

    Simple . . . the book is wrong!

    I'm puzzled . . . you got the same wrong answer.
    . . Did you just copy the book's answer and trust it to be correct?

    Are you aware that the derivative of \arctan x is: . \frac{1}{1+x^2}

    There's no way to get \sin^2x in the denominator
    . . and you should know that.

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  5. #5
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    Hello

    of course not I don't trust my book but the problem is if I integrate it by derive on my computer he tells me that integrate of \frac{ln(1+sin(x))}{sin^2(x)}

    will be ln(tan \frac{x}{2})-cot(x)ln(sin(x)+1)-x

    here a screen shot of my window

    Greets.
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  6. #6
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    Hello, Bert!

    The problem is if I integrate it by derive on my computer

    it tells me that: . \int\frac{\ln(1+ \sin x )}{\sin^2x}\,dx \;= \;\ln\left(\tan\frac{x}{2}\right) - \underbrace{\cot x}\cdot\ln(\sin x +1) -x

    I don't see a problem . . .
    That is not the original function: . \arctan x\cdot\ln(1 + \sin x)

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  7. #7
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    maybe the problem is that we define in an other way cotan I mean \frac{1}{tan}

    you use arctan I think thats the function who give back an angle I don't use this.

    Maybe I copy the exercise at a wrong way so let me show an excat copy.



    Greets Tanks a lot.
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  8. #8
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    Hello, Bert!

    Sheesh! . . . That explains our different answers . . .



    The function is: . y \;= \;\ln\left(\tan \frac{x}{2}\right) - \underbrace{\cot x}\cdot\ln(1 + \sin x) - x

    Your notation was incorrect and misleading . . .


    We know that: . \tan^3x means (\tan x)^3

    . . but \cot x \:=\:\frac{1}{\tan x} and should be written (\tan x)^{-1} **


    The symbols . \arctan x,\;\tan^{-1}x\text{, and your {\bf atan} }x
    . . are used to signify the inverse tangent of x

    That is, \text{asin }x does not mean \frac{1}{\sin x}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **
    I don't understand why you replaced \cot x with its reciprocal.
    . . There is a formula: . (\cot x)' \;= \; -\csc^2x

    Last edited by Soroban; July 5th 2006 at 08:36 AM.
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  9. #9
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    this will be the problem I think The matter is that ther are different notations in different language area.

    So let we work with the original copy of my book.

    Where do I miss then in my answer?

    Greets Thanks.
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  10. #10
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    Hello again, Bert!

    I got it! . . .
    They did more simplifying than I thought humanly possible!


    We have: . y \;= \;\ln\left(\tan\frac{x}{2}\right) - \cot x\cdot\ln(1 + \sin x) - x


    Then: . y' \;= \;\frac{1}{2}\cdot\frac{\sec^2\frac{x}{2}}{\tan \frac{x}{2}} -  \left[\cot x\cdot\frac{\cos x}{1 + \sin x} - \csc^2x\cdot\ln(1 + \sin x)\right] - 1


    I've already shown that the first fraction can be simplfied.

    . . So we have: . y' \;= \;\csc x - \frac{\cot x\cdot\cos x}{1 + \sin x} + \csc^2x\cdot\ln(1 + \sin x) - 1


    The second fraction is: . \frac{\frac{\cos x}{\sin x}\cdot\cos x}{1 + \sin x} \;= \;\frac{\cos^2x}{\sin x(1 + \sin x)}\;= \;\frac{1 - \sin^2x}{\sin x(1 + \sin x)}

    . . . =\;\frac{(1 - \sin x)(1 + \sin x)}{\sin x(1 + \sin x)} \;= \;\frac{1 - \sin x}{\sin x}\;= \frac{1}{\sin x} - \frac{\sin x}{\sin x}\;=\;\csc x - 1


    The third term is: . \csc^2x\cdot\ln(1 + \sin x)\;=\;\frac{\ln(1 + \sin x)}{\sin^2x}


    . . So we have: . y' \;= \;\csc x -(\csc x - 1) + \frac{\ln(1 + \sin x)}{\sin^2x} - 1


    Therefore: . \boxed{y' \;= \;\frac{\ln(1 + \sin x)}{\sin^2x}} . . . ta-DAA!

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  11. #11
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    thank you very much I see.

    Greets.
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