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To determine \( q'(x) \) for the given function:

\[
q(x) = \frac{8(4^x)}{x^4},
\]

we need to apply the quotient rule for derivatives. The quotient rule states that for a function \( \frac{f(x)}{g(x)} \), the derivative is:

\[
\left( \frac{f(x)}{g(x)} \right)' = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}.
\]

Here, \( f(x) = 8(4^x) \) and \( g(x) = x^4 \).

### Step 1: Derivatives of \( f(x) \) and \( g(x) \)
1. \( f(x) = 8(4^x) \).
   - The derivative of \( 4^x \) is \( 4^x \ln(4) \), so:
   \[
   f'(x) = 8(4^x)\ln(4).
   \]

2. \( g(x) = x^4 \).
   - The derivative is:
   \[
   g'(x) = 4x^3.
   \]

### Step 2: Applying the Quotient Rule
Now apply the quotient rule:
\[
q'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}.
\]
Substitute the expressions for \( f(x) \), \( f'(x) \), \( g(x) \), and \( g'(x) \):

\[
q'(x) = \frac{8(4^x)(\ln(4))(x^4) - 8(4^x)(4x^3)}{(x^4)^2}.
\]

### Step 3: Simplification
Factor out \( 8(4^x) \):

\[
q'(x) = \frac{8(4^x) \left( (\ln(4))(x^4) - 4x^3 \right)}{(x^4)^2}.
\]

Now, comparing this result to the provided choices in the image, the correct answer is the first option:

\[
q'(x) = \frac{8(4^x)(4x^3) - 8(4^x)(\ln(4))(x^4)}{(x^4)^2}.
\]

---

Do you want more details on any part of this solution? Here are five related questions:

1. How is the derivative of \( 4^x \) computed?
2. What are some other examples of the quotient rule in action?
3. Could we simplify \( q(x) \) further before differentiating?
4. What if the function was \( \frac{8(4^x)}{x^3} \), how would that change the derivative?
5. What is the significance of \( \ln(4) \) in the derivative process?

**Tip:** When applying the quotient rule, ensure both the numerator and denominator derivatives are calculated correctly before simplifying the final result.

Determine q'(x) when q(x) = \frac{8(4^x)}{x^4}.

This solution explains how to find the derivative of q(x) = 8(4^x)/x^4 using the quotient rule. Key calculus concepts include the quotient rule for derivatives and the differentiation of exponential and power functions. Suitable for students in Grades 10-12.

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