To differentiate $f(t) = \frac{2 + \ln t}{2 - \ln t}$, we use the quotient rule, which states:

$\frac{d}{dt} \left( \frac{u(t)}{v(t)} \right) = \frac{u'(t)v(t) - u(t)v'(t)}{(v(t))^2}.$

Here, $u(t) = 2 + \ln t$ and $v(t) = 2 - \ln t$.

Step 1: Differentiate $u(t)$ and $v(t)$

• $u(t) = 2 + \ln t \implies u'(t) = \frac{1}{t}$
• $v(t) = 2 - \ln t \implies v'(t) = -\frac{1}{t}$

Step 2: Apply the quotient rule

Substitute $u(t), v(t), u'(t), v'(t)$ into the formula:

$f'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{(v(t))^2}.$

$f'(t) = \frac{\left(\frac{1}{t}\right)(2 - \ln t) - (2 + \ln t)\left(-\frac{1}{t}\right)}{(2 - \ln t)^2}.$

Step 3: Simplify the numerator

Expand the terms in the numerator:

$f'(t) = \frac{\frac{2 - \ln t}{t} + \frac{2 + \ln t}{t}}{(2 - \ln t)^2}.$

Combine the terms:

$f'(t) = \frac{\frac{2 - \ln t + 2 + \ln t}{t}}{(2 - \ln t)^2}.$

$f'(t) = \frac{\frac{4}{t}}{(2 - \ln t)^2}.$

Final Answer:

$f'(t) = \frac{4}{t(2 - \ln t)^2}.$

Let me know if you'd like a deeper breakdown or have further questions! Here are some related queries:

1. How does the quotient rule work in general cases?
2. Can you differentiate other logarithmic functions like $\ln(ax)$ or $\ln(x^n)$?
3. What are the properties of logarithmic differentiation?
4. How do we simplify more complex rational derivatives?
5. Why is the chain rule not directly applicable here?

Tip: Always check if functions can be simplified before differentiating—it might make the process easier.