The given problem asks to find the derivative of $y$ with respect to $t$, where:

$y = \ln(t^6)$

Solution:

First, we can simplify $\ln(t^6)$ using the logarithmic property:

$\ln(a^b) = b \cdot \ln(a)$

Thus:

$y = 6 \ln(t)$

Now, differentiate $y = 6 \ln(t)$ with respect to $t$:

$\frac{dy}{dt} = 6 \cdot \frac{d}{dt}[\ln(t)]$

The derivative of $\ln(t)$ is $\frac{1}{t}$, so:

$\frac{dy}{dt} = 6 \cdot \frac{1}{t} = \frac{6}{t}$

Final Answer:

$\frac{dy}{dt} = \frac{6}{t}$

Let me know if you would like more details or have any questions.

Here are five related questions to explore further:

1. How would the derivative change if $y = \ln(t^n)$ for any $n$?
2. What is the second derivative of $y = \ln(t^6)$?
3. How does the chain rule apply to logarithmic functions?
4. What happens to the derivative if $t$ is replaced by another function, like $t^2 + 1$?
5. How can the product rule be used in problems involving logarithms?

Tip: Always remember to apply the logarithmic properties to simplify expressions before differentiating, as it can make the process much easier.