- #1
karush
Gold Member
MHB
- 3,240
- 5
$\tiny{242.2q.3}$
$\textsf{Find the derivative}\\$
\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}
$\textit{think this is ok, but suggestions before I cp it into overleaf?}$
$\textsf{Find the derivative}\\$
\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}
$\textit{think this is ok, but suggestions before I cp it into overleaf?}$
