- #1

karush

Gold Member

MHB

- 3,269

- 5

$\textsf{Find the derivative}\\$

\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}

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