To find the derivative of $h(x) = 4^{(x^9)}$, we can use a combination of the chain rule and the exponential derivative rule.

Step-by-Step Explanation:

1. Recall the derivative of $a^{u(x)}$: The derivative of $a^{u(x)}$ with respect to $x$, where $a$ is a constant, is: $\frac{d}{dx} \left( a^{u(x)} \right) = a^{u(x)} \cdot \ln(a) \cdot \frac{d}{dx}[u(x)]$ Here, $a = 4$ and $u(x) = x^9$.

2. Apply the derivative to $4^{x^9}$: Using the formula above, we have: $h'(x) = 4^{x^9} \cdot \ln(4) \cdot \frac{d}{dx}(x^9)$

3. Differentiate $x^9$: The derivative of $x^9$ is: $\frac{d}{dx}(x^9) = 9x^8$

4. Combine everything: Substituting the derivative of $x^9$ into the equation, we get: $h'(x) = 4^{x^9} \cdot \ln(4) \cdot 9x^8$

Thus, the first derivative of $h(x) = 4^{(x^9)}$ is: $h'(x) = 4^{x^9} \cdot 9x^8 \cdot \ln(4)$

Correct Answer:

The correct answer from the options provided is: $h'(x) = 4^{x^9} \cdot 9x^8 \cdot (\ln 4)$ This corresponds to the third option: $h'(x) = 4^{x^9} \cdot 9x^8 \cdot \ln(4)$.

Would you like further clarification on any part of the process? Here are 5 related questions you might find helpful:

1. How does the chain rule apply when differentiating composite functions?
2. What is the general form of the derivative of exponential functions?
3. How do you handle differentiation of powers like $x^9$?
4. Can this method be extended to functions with bases other than 4?
5. How is the natural logarithm $\ln(a)$ important in differentiating exponential functions?

Tip: Always remember that when differentiating exponential functions with a variable exponent, both the base and the exponent need to be considered in the derivative process.