We are given the function $f(x) = \ln(3 + e^{8x})$ and asked to find $f'(0)$, the derivative of $f(x)$ evaluated at $x = 0$.

Step 1: Find the derivative of $f(x)$.

To differentiate $f(x) = \ln(3 + e^{8x})$, we'll use the chain rule and the derivative of the natural logarithm:

$\frac{d}{dx} \ln(g(x)) = \frac{g'(x)}{g(x)}$

Here, $g(x) = 3 + e^{8x}$. Now, we differentiate $g(x)$:

$g'(x) = \frac{d}{dx}(3 + e^{8x}) = 0 + 8e^{8x} = 8e^{8x}$

Using the chain rule, the derivative of $f(x)$ is:

$f'(x) = \frac{g'(x)}{g(x)} = \frac{8e^{8x}}{3 + e^{8x}}$

Step 2: Evaluate $f'(x)$ at $x = 0$.

Substitute $x = 0$ into the derivative:

$f'(0) = \frac{8e^{8(0)}}{3 + e^{8(0)}} = \frac{8e^0}{3 + e^0} = \frac{8 \times 1}{3 + 1} = \frac{8}{4} = 2$

Final Answer:

$f'(0) = 2$

Would you like further details on any part of the solution?

Additional Questions:

1. What is the general rule for differentiating logarithmic functions?
2. How would the solution change if the exponent in $e^{8x}$ were different?
3. Can you explain the chain rule in more detail with another example?
4. How can you apply this differentiation to other exponential functions?
5. What is the significance of evaluating a derivative at a specific point?

Tip:

Always remember that the chain rule is your best tool when differentiating composite functions like $\ln(3 + e^{8x})$!