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

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The derivative of the function y = (1 + ln(t)) / (1 - ln(t)) is calculated using the quotient rule. The derivatives of the numerator and denominator are f' = 1/t and g' = 1/t, respectively. Applying the quotient rule results in y' = 2 / (t(ln(t) - 1)^2). The calculation is confirmed to be correct, with one user expressing satisfaction with the result. Overall, the discussion focuses on verifying the derivative and ensuring accuracy before finalizing the work.
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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)
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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