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Thread: Differentiation of inverse tanh

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    Differentiation of inverse tanh

    inverse tanh(X) = 1/2 ln(1+x/1-x) differentiate to show that d/dx = 1/1-x^2

    I have been working on this problem for so long, can anyone show me the solution. Thanks!
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  2. #2
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    Quote Originally Posted by skeltonjoe View Post
    inverse tanh(X) = 1/2 ln(1+x/1-x) differentiate to show that d/dx = 1/1-x^2

    I have been working on this problem for so long, can anyone show me the solution. Thanks!
    $\displaystyle \tanh^{-1}{x} = \frac{1}{2}\ln{\frac{1 + x}{1 - x}}$.


    So if $\displaystyle y = \tanh^{-1}{x}$

    $\displaystyle y = \frac{1}{2}\ln{\frac{1 + x}{1 - x}}$

    Let $\displaystyle u = \frac{1 + x}{1 - x}$ so that $\displaystyle y = \frac{1}{2}\ln{u}$.


    $\displaystyle \frac{du}{dx} = \frac{(1 - x)\frac{d}{dx}(1 + x) - (1 + x)\frac{d}{dx}(1 - x)}{(1 - x)^2}$ by the Quotient Rule

    $\displaystyle = \frac{1 - x + 1 + x}{(1 - x)^2}$

    $\displaystyle = \frac{2}{(1 - x)^2}$.


    $\displaystyle \frac{dy}{du} = \frac{1}{2}\cdot \frac{1}{u}$

    $\displaystyle = \frac{1}{2}\cdot \frac{1}{\frac{1 + x}{1- x}}$

    $\displaystyle = \frac{1}{2}\cdot \frac{1 - x}{1 + x}$

    $\displaystyle = \frac{1 - x}{2(1 + x)}$.


    Therefore

    $\displaystyle \frac{dy}{dx} = \frac{2}{(1 - x)^2}\cdot \frac{1 - x}{2(1 + x)}$

    $\displaystyle =\frac{1}{(1 - x)(1 + x)}$

    $\displaystyle = \frac{1}{1 - x^2}$.
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  3. #3
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    Quote Originally Posted by skeltonjoe View Post
    inverse tanh(X) = 1/2 ln(1+x/1-x) differentiate to show that d/dx = 1/1-x^2

    I have been working on this problem for so long, can anyone show me the solution. Thanks!
    using log properties

    $\displaystyle y = tanh(X) = \frac{1}{2} \ln{\frac{1+x}{1-x}} = \frac{1}{2}(\ln{(1+x)} - \ln{(1-x)})$

    so

    $\displaystyle y' = \frac{1}{2}\bigg(\frac{1}{1+x} +\frac{1}{1-x}\bigg) = \frac{1}{2}\bigg(\frac{2}{1-x^2}\bigg) = \frac{1}{1-x^2}$
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