242.2q.3 Find the derivative (1+ln{(t))/(1-ln{(t))

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The derivative of the function \( y = \frac{1+\ln{(t)}}{1-\ln{(t)}} \) is accurately calculated as \( y' = \frac{2}{t(\ln(t)-1)^2} \). The calculation involves applying the quotient rule, where \( f = 1+\ln(t) \) and \( g = \ln(t)-1 \). The derivatives \( f' \) and \( g' \) are both \( \frac{1}{t} \). The final expression simplifies correctly, confirming the accuracy of the derivative.

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$\tiny{242.2q.3}$
$\textsf{Find the derivative}\\$
\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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The derivative is correct! (Yes)
 
took me too long to do it...(Wasntme)
 

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