To find $f'(t)$, we need to differentiate the given function:

$f(t) = 5 \ln(t) - 3e^{-2t}$

Step-by-step differentiation:

1. Differentiate $5 \ln(t)$:

   The derivative of $\ln(t)$ is $\frac{1}{t}$. So: $\frac{d}{dt} \left( 5 \ln(t) \right) = \frac{5}{t}$

2. Differentiate $-3e^{-2t}$:

   The derivative of $e^{u(t)}$ where $u(t)$ is a function of $t$, is $e^{u(t)} \cdot u'(t)$. Here, $u(t) = -2t$, so: $\frac{d}{dt} \left( -3e^{-2t} \right) = -3 \cdot e^{-2t} \cdot (-2) = 6e^{-2t}$

Final result:

Now, putting the two derivatives together:

$f'(t) = \frac{5}{t} + 6e^{-2t}$

Would you like more details or have any questions?

Here are 5 related questions for deeper exploration:

1. How do you find the second derivative $f''(t)$ of this function?
2. How does the natural logarithm $\ln(t)$ behave as $t \to 0^+$?
3. What is the behavior of $f(t)$ as $t \to \infty$?
4. How do you find the critical points of the function $f(t)$?
5. What does the graph of $f'(t)$ look like?

Tip: When differentiating exponential functions, always pay attention to the chain rule if the exponent involves a variable.