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Math Help - hyperbolic differentiation

  1. #1
    Member i_zz_y_ill's Avatar
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    hyperbolic differentiation

    Q (i)prove that d/dx(artanhx)=1/(1-x^2)

    this is just artanhx=y, tanhy=x then diff.

    can't do simplify end of (ii) Q by using partial ractions and integrating deduce from this(ans to (i)) the logarithmic form of artanhx,,,i know the log form of artanhx is 1/2ln(1+x/1-x)???????

    i did 1/(1-x^2) = (int)-1/(-2x-2) -(int)1/(2x-2)

    giving 1/2ln|-2x-2| -ln|2x-2| then i get to 1/2ln|(-2x-2)/(2x-2)| not sure where to go from here i think its easy but im being ignorant,,,the log form of artanhx is 1/2ln(1+x/1-x)???????pls help!!!!
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  2. #2
    Eater of Worlds
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    Let's start by proving that tanh^{-1}(x)=\frac{1}{2}ln(\frac{1+x}{1-x})

    Let y=tanh^{-1}(x), \;\ x=tanh(y)=\frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}=\frac{e^{2y}-1}{e^{2y}+1}

    xe^{2y}+x=e^{2y}-1

    e^{2y}(x-1)=-x-1,

    e^{2y}=\frac{1+x}{1-x}, \;\ 2y=ln(\frac{1+x}{1-x}), \;\ y=\frac{1}{2}ln(\frac{x+1}{x-1})

    Now, \frac{d}{dx}[tanh^{-1}(x)]=\frac{d}{dx}[\frac{1}{2}(ln(1+x)-ln(1-x))]=\frac{1}{2}\left(\frac{1}{1+x}+\frac{1}{1-x}\right)=\frac{1}{1-x^{2}}
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  3. #3
    Member i_zz_y_ill's Avatar
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    cheers

    i used (-x-1)(x-1) for the 'base' of my partial fractions,,, shudda used (x+1)(-x+1) cheers fella!
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  4. #4
    Member i_zz_y_ill's Avatar
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    :o

    oh no wait when i use either the partial expantion of (x+1)(-x+1) or the base of (-x-1)(x-1) i still cant expand to get this,,
    with (x+1)(-x+1) i get 1/2ln(2x+2)+l(x-1)

    and with (-x-1)(x-1) i get 1/2ln(-2x-2)-1/2ln(2x-2) u know where to go from here??
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  5. #5
    Eater of Worlds
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    They are the same:

    \frac{-x-1}{x-1}=\frac{1+x}{1-x}
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