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  1. MHB Master
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    #1
    $\tiny{242.2q.3}$
    $\textsf{Find the derivative}\\$
    \begin{align}
    \displaystyle
    y&=\frac{1+\ln{(t)}}{1-\ln{(t)}}
    =-\frac{1+\ln(t)}{\ln(t)-1}=\frac{f}{g}\\
    f&=1+\ln(t) \therefore f'=\frac{1}{t}\\
    g&=\ln(t)-1 \therefore g'=\frac{1}{t}\\
    y'&= \frac{f\cdot g' - f'\cdot g}{g^2}\\
    &=\frac{(1+\ln(t))(1/t)-(\ln(t)-1)(1/t)}{(\ln(t)-1)^2} \\
    &=\frac{1+\ln(t)-\ln(t)+1}{t(\ln(t)-1)^2}\\
    &=\dfrac{2}{t\left(\ln\left(t\right)-1\right)^2}
    \end{align}

    $\textit{think this is ok, but suggestions before I cp it into overleaf???}$

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    #2
    The derivative is correct!!

  3. MHB Master
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    #3 Thread Author
    took me too long to do it...

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