To differentiate the function $g(t) = \ln(6 + t^8)$, we will use the chain rule.

Steps:

1. Differentiate the outer function: The derivative of $\ln(x)$ with respect to $x$ is $\frac{1}{x}$. So, for $g(t) = \ln(6 + t^8)$, treat $6 + t^8$ as the inner function. $\frac{d}{dt} \left( \ln(6 + t^8) \right) = \frac{1}{6 + t^8} \cdot \frac{d}{dt}(6 + t^8)$

2. Differentiate the inner function: The derivative of $6 + t^8$ with respect to $t$ is $8t^7$ (since $6$ is constant, its derivative is 0). $\frac{d}{dt}(6 + t^8) = 8t^7$

3. Combine the results: Multiply the derivatives of the outer and inner functions. $g'(t) = \frac{1}{6 + t^8} \cdot 8t^7 = \frac{8t^7}{6 + t^8}$

Thus, the derivative of $g(t) = \ln(6 + t^8)$ is: $g'(t) = \frac{8t^7}{6 + t^8}$

Would you like further details or have any questions?

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Here are 5 related questions to explore:

1. What is the derivative of $g(t) = \ln(t^8 + t^2)$?
2. How do you apply the chain rule in general cases of logarithmic functions?
3. What happens to the derivative as $t \to 0$ or $t \to \infty$ in the function $g(t)$?
4. How would you find the second derivative of $g(t) = \ln(6 + t^8)$?
5. What is the derivative of $g(t) = \ln(a + t^n)$ where $a$ and $n$ are constants?

Tip: When differentiating composite functions like $\ln(f(t))$, always apply the chain rule by differentiating the outer and inner functions separately.