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We are given the function \( g(x) = \arcsin(e^{4x}) + 3 \), and the task is to find the derivative of \( g(x) \).

### Step-by-Step Solution:

1. The derivative of the constant \( 3 \) is \( 0 \), so we can focus on differentiating \( \arcsin(e^{4x}) \).

2. The derivative of \( \arcsin(u) \) with respect to \( x \) is given by:
   \[
   \frac{d}{dx} [\arcsin(u)] = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}
   \]
   where \( u = e^{4x} \) in this case.

3. Now, calculate the derivative of \( u = e^{4x} \):
   \[
   \frac{du}{dx} = 4e^{4x}
   \]

4. Applying the chain rule, we get:
   \[
   \frac{d}{dx} \arcsin(e^{4x}) = \frac{1}{\sqrt{1 - (e^{4x})^2}} \cdot 4e^{4x}
   \]

5. Simplify the expression inside the square root:
   \[
   (e^{4x})^2 = e^{8x}
   \]
   So, the derivative becomes:
   \[
   \frac{d}{dx} g(x) = \frac{4e^{4x}}{\sqrt{1 - e^{8x}}}
   \]

Thus, the derivative of \( g(x) \) is:
\[
g'(x) = \frac{4e^{4x}}{\sqrt{1 - e^{8x}}}
\]

Let me know if you need any further details!

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Here are five related questions for you to consider:
1. How does the derivative of the \( \arcsin \) function change if the argument involves different powers of \( e^x \)?
2. Can you compute the second derivative of the function \( g(x) \)?
3. What is the derivative of \( g(x) = \arcsin(e^{4x}) + 3 \) evaluated at \( x = 0 \)?
4. How would you find the critical points of the function \( g(x) \)?
5. How does the behavior of the function \( g(x) \) change as \( x \to \infty \)?

**Tip**: When differentiating composite functions like \( \arcsin(f(x)) \), always remember to use the chain rule effectively!

What is the derivative of g(x) = arcsin(e^(4x)) + 3?

Learn how to find the derivative of g(x) = arcsin(e^(4x)) + 3. This problem covers differentiation techniques using the chain rule and the derivative of inverse trigonometric functions. Detailed step-by-step solution included. Suitable for advanced high school or university-level calculus.

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