To solve for $f'(t)$ given that

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

we'll differentiate each term separately with respect to $t$.

Step 1: Differentiate $2 \ln(t)$

The derivative of $\ln(t)$ with respect to $t$ is $\frac{1}{t}$. So,

$\frac{d}{dt} \left( 2 \ln(t) \right) = 2 \cdot \frac{1}{t} = \frac{2}{t}.$

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

Using the chain rule, the derivative of $e^{-3t}$ is $-3 e^{-3t}$. Thus,

$\frac{d}{dt} \left( 2 e^{-3t} \right) = 2 \cdot (-3 e^{-3t}) = -6 e^{-3t}.$

Combine Results

Now, add the derivatives from each part:

$f'(t) = \frac{2}{t} - 6 e^{-3t}.$

Conclusion

The correct answer is:

$\frac{2}{t} - 6 e^{-3t}$

This matches the third option: $\frac{2}{t} - 6 e^{-3t}$.

Would you like further details, or do you have any questions about the solution?

Related Questions

1. How do you differentiate logarithmic functions with coefficients?
2. What is the chain rule and how is it applied to exponential functions?
3. How do you approach the differentiation of functions with mixed terms, like logarithmic and exponential?
4. Why does the derivative of $e^{at}$ include a multiplier of $a$?
5. How do derivatives of logarithmic functions behave near $t = 0$?

Tip

When differentiating functions with constants multiplied by terms, you can factor out the constants first, then apply the derivative to the term inside. This simplifies the process.