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Math Help - Proof of derivative

  1. #1
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    Proof of derivative

    I'm stuck on this question. How would I go about doing it? Thanks in advance.

    Q: Prove that the derivative of \mathrm{artanh} x, -1 < x < 1, is \frac{1}{1-x^2}.
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    Start by showing that \text{arctanh }x=\frac{1}{2}\ln \frac{1+x}{1-x}. After that contemplate its derivative.
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  3. #3
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    Hello,

    You can do this way :

    y=\tanh(x)=\frac{\sinh(x)}{\cosh(x)}=\frac{e^x-e^{-x}}{2} \cdot \frac{2}{e^x+e^{-x}}= \boxed{\frac{e^x-e^{-x}}{e^x+e^{-x}}}

    x=\tanh^{-1}(y)

    So we have to find the inverse of that.

    y=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^x}{e^x} \cdot \frac{e^x-e^{-x}}{e^x+e^{-x}}

    y=\frac{e^{2x}-1}{e^{2x}+1}

    Let X=e^{2x} (easier to type )


    y=\frac{X-1}{X+1}

    --> y(X+1)=X-1

    ----> X(y-1)=-1-y

    -------> \boxed{X=\frac{1+y}{1-y}}


    e^{2x}=\frac{1+y}{1-y}

    ---> 2x=\ln(1+y)-\ln(1-y)

    ------> \boxed{\color{red} x=\tanh^{-1}(y)=\frac 12 \left(\ln(1+y)-\ln(1-y)\right)}




    Then derivate this...
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  4. #4
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    Sorry, my mind is currently completely blank.

    \frac{\mathrm{d} (\mathrm{ln}x)}{\mathrm{d}x} = \frac{1}{x}

    Hence doesn't the derivative of \frac{1}{2}\ln \frac{1+x}{1-x} = 1 / (\frac{1+x}{1-x}) = \frac{1-x}{1+x}
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  5. #5
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    Quote Originally Posted by Air View Post
    Sorry, my mind is currently completely blank.

    \frac{\mathrm{d} (\mathrm{ln}x)}{\mathrm{d}x} = \frac{1}{x}

    Hence doesn't the derivative of \frac{1}{2}\ln \frac{1+x}{1-x} = 1 / (\frac{1+x}{1-x}) = \frac{1-x}{1+x}
    Don't forget chain rule

    \left(\frac{1}{2}\ln \frac{1+x}{1-x}\right)' = \frac1{\left(\frac{1+x}{1-x}\right)}\frac{d}{dx}\left(\frac{1+x}{1-x}\right)
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  6. #6
    Moo
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    Transform : \ln \frac{1+x}{1-x}=\ln(1+x)-\ln(1-x)

    I guess it'd be easier this way
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  7. #7
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    Got so close but have made an error in my algebra (I think). Can you spot where I have gone wrong.


    \frac{1}{2}\ln \frac{1+x}{1-x} = \frac{1}{2} \ln (1+x) - \frac{1}{2} \ln (1-x)


    Derivative:
    \frac{1}{2}.\frac{1}{1+x} - \frac{1}{2}.\frac{1}{1-x}

    \frac{1}{2+2x} - \frac{1}{2-2x}

    \frac{(2-2x) - (2+2x)}{(2+2x)(2-2x)}

    \frac{-4x}{-4x^2 +4}

    \frac{-4x}{4(1-x^2)}

    \frac{-x}{1-x^2}
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  8. #8
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    Quote Originally Posted by Air View Post

    \frac{1}{2}\ln \frac{1+x}{1-x} = \frac{1}{2} \ln (1+x) - \frac{1}{2} \ln (1-x)


    Derivative:
    \frac{1}{2}.\frac{1}{1+x} - \frac{1}{2}.\frac{1}{1-x}
    You missed the chain rule by takin' the derivative of \ln(1-x).

    (Typo fixed.)
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  9. #9
    Moo
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    The derivative of \ln(1-x) is {\color{red}-} \frac{1}{1-x}, since the derivative of 1-x is -1
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  10. #10
    Moo
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    Quote Originally Posted by Krizalid View Post
    You missed the chain rule by takin' the derivative of \frac1{1-x}.
    Of \ln(1-x), right ?
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  11. #11
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    Hi
    How about using \frac{\mathrm{d}}{\mathrm{d}x}\left(f^{-1}(x)\right)=\frac{1}{f'\circ f^{-1}(x)} with f(x)=\tanh x,\,f^{-1}(x)=\tanh^{-1}x,\,f'(x)=\tanh' x=1-\tanh^2x ?

    If you don't know this, it comes from the derivative of f\circ f^{-1}(x)=x.
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