To evaluate the derivative of the function $f(x) = (2x)^{4x}$, we will use logarithmic differentiation since we have a function in the form $g(x)^{h(x)}$, where both the base and the exponent depend on $x$.

Step-by-step solution:

1. Rewrite the function: $f(x) = (2x)^{4x}$

2. Take the natural logarithm of both sides: $\ln(f(x)) = \ln\left( (2x)^{4x} \right)$ Using the property of logarithms $\ln(a^b) = b \ln(a)$, this becomes: $\ln(f(x)) = 4x \ln(2x)$

3. Differentiate both sides implicitly: On the left side, we use the chain rule: $\frac{d}{dx} \left( \ln(f(x)) \right) = \frac{f'(x)}{f(x)}$

   On the right side, we use the product rule to differentiate $4x \ln(2x)$. Recall that $\ln(2x) = \ln(2) + \ln(x)$, so: $\frac{d}{dx} \left( 4x \ln(2x) \right) = \frac{d}{dx} \left( 4x (\ln(2) + \ln(x)) \right)$ $= 4 \ln(2x) + 4x \cdot \frac{1}{x} = 4 \ln(2x) + 4$

4. Solve for $f'(x)$: Now, we have: $\frac{f'(x)}{f(x)} = 4 \ln(2x) + 4$ Multiply both sides by $f(x) = (2x)^{4x}$: $f'(x) = (2x)^{4x} \cdot (4 \ln(2x) + 4)$

Final Answer:

$f'(x) = (2x)^{4x} \cdot (4 \ln(2x) + 4)$

This is the derivative of the function $f(x) = (2x)^{4x}$.

---

Would you like a more detailed explanation of any step? Here are 5 related questions to expand your understanding:

1. How do we apply logarithmic differentiation in general?
2. What is the product rule and how is it applied in this problem?
3. Why do we need to use the chain rule when differentiating $\ln(f(x))$?
4. How do we differentiate a function that involves both a variable base and a variable exponent?
5. How does the logarithmic property $\ln(a^b) = b \ln(a)$ work in other examples?

Tip: Logarithmic differentiation is especially useful when dealing with functions where both the base and exponent are variable, as it simplifies the differentiation process significantly.