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To solve the problem of differentiating \((6 + F(2x))^6\) with respect to \(x\), we'll use the **chain rule**. Here’s the step-by-step solution:

### Step 1: Identify the Outer and Inner Functions
Let:
\[
u = 6 + F(2x)
\]
and thus we are differentiating \(u^6\) with respect to \(x\).

### Step 2: Apply the Chain Rule
The derivative of \(u^6\) with respect to \(x\) is:
\[
\frac{d}{dx} \left( u^6 \right) = 6u^5 \cdot \frac{du}{dx}
\]

### Step 3: Differentiate \(u = 6 + F(2x)\)
Now, find \(\frac{du}{dx}\):
\[
\frac{du}{dx} = \frac{d}{dx} \left( 6 + F(2x) \right) = F'(2x) \cdot 2
\]
since the derivative of \(F(2x)\) with respect to \(x\) is \(F'(2x) \cdot 2\).

### Step 4: Substitute \(u\) and \(\frac{du}{dx}\)
Now, substitute back \(u = 6 + F(2x)\) and \(\frac{du}{dx} = 2F'(2x)\) into the chain rule expression:
\[
\frac{d}{dx} \left( (6 + F(2x))^6 \right) = 6(6 + F(2x))^5 \cdot 2F'(2x)
\]

### Step 5: Simplify the Expression
\[
= 12(6 + F(2x))^5 F'(2x)
\]

### Conclusion
The correct answer is therefore:
**A. \(12(6 + F(2x))^5 F'(2x)\)**

Would you like further explanation on any step?

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Here are 5 additional questions to deepen understanding:

1. How would the result change if the exponent were different, say \((6 + F(2x))^4\)?
2. What would happen if \(F(x)\) were not differentiable at some points?
3. How does the chain rule apply when there are more than two layers of functions?
4. Could we use any other differentiation technique for this problem besides the chain rule?
5. What is the significance of \(F'(2x)\) in the context of this problem?

**Tip:** When using the chain rule, always identify inner and outer functions clearly to simplify the differentiation process.

Express the indicated derivative in terms of F(x). Assume that F is differentiable. Find d/dx (6 + F(2x))^6.

Learn how to find the derivative of (6 + F(2x))^6 with respect to x using the chain rule. This problem covers key calculus concepts, including differentiation and the application of the chain rule for composite functions. Suitable for high school calculus students.

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