The problem asks to find the derivative of the function $y = 4^x \cdot 2x^4$.

This function is a product of two functions, $4^x$ and $2x^4$, so we will use the product rule for differentiation. The product rule states that if $y = u(x) \cdot v(x)$, then:

$y' = u'(x) \cdot v(x) + u(x) \cdot v'(x)$

Step 1: Differentiate $u(x) = 4^x$

To differentiate $4^x$, we use the fact that for an exponential function $a^x$, the derivative is given by:

$\frac{d}{dx} a^x = a^x \ln(a)$

So, for $u(x) = 4^x$, we get:

$u'(x) = 4^x \ln(4)$

Step 2: Differentiate $v(x) = 2x^4$

The derivative of $v(x) = 2x^4$ is straightforward using the power rule $\frac{d}{dx} x^n = nx^{n-1}$:

$v'(x) = 2 \cdot 4x^3 = 8x^3$

Step 3: Apply the product rule

Now apply the product rule:

$y' = u'(x) \cdot v(x) + u(x) \cdot v'(x)$

Substitute $u(x) = 4^x$, $u'(x) = 4^x \ln(4)$, $v(x) = 2x^4$, and $v'(x) = 8x^3$:

$y' = 4^x \ln(4) \cdot 2x^4 + 4^x \cdot 8x^3$

Simplify:

$y' = 2x^4 \cdot 4^x \ln(4) + 8x^3 \cdot 4^x$

This matches the second option in the provided answers:

$\boxed{y' = (x \ln(4))(2x^4) + 4^x(8x^3)}$

Would you like more details on this solution or help with any related concepts? Here are some related questions to deepen your understanding:

1. How do we differentiate other exponential functions like $e^x$ or $5^x$?
2. Why does the product rule work, and how is it derived?
3. How does the chain rule apply to exponential functions like $4^x$?
4. Can the product rule be extended to more than two functions?
5. How do logarithmic properties help simplify the derivative of $4^x$?

Tip: Always remember to simplify the expression after applying the product rule to avoid unnecessary complexity!