To find the derivative $f'(x)$ of the function $f(x) = 4 \cdot \arcsin(e^{8x})$, we will apply the chain rule.

1. Differentiate the outer function: The derivative of $\arcsin(u)$ is $\frac{1}{\sqrt{1 - u^2}}$ where $u = e^{8x}$.

2. Differentiate the inner function: The derivative of $e^{8x}$ is $8e^{8x}$.

Now, using the chain rule:

$f'(x) = 4 \cdot \frac{1}{\sqrt{1 - (e^{8x})^2}} \cdot (8e^{8x})$

Thus, we can write:

$f'(x) = \frac{32 e^{8x}}{\sqrt{1 - e^{16x}}}$

This is the derivative of the function.

Summary

$f'(x) = \frac{32 e^{8x}}{\sqrt{1 - e^{16x}}}$

Would you like to see a detailed explanation of any steps? Here are some follow-up questions:

1. What is the significance of the chain rule in calculus?
2. Can you explain the properties of the arcsine function?
3. How do you evaluate the derivative at a specific point?
4. What happens to $f'(x)$ as $x$ approaches certain limits?
5. Can you give examples of other functions involving arcsine?

Tip: Always check if the inner function is within the domain of the arcsine to avoid undefined values.