Let's differentiate the function $y = x^2 (1 - 4x)^4$ using the product rule and the chain rule.

Steps:

1. Product Rule: The product rule states that if we have a function of the form $y = u(x) v(x)$, the derivative is given by:

   $\frac{dy}{dx} = u'(x) v(x) + u(x) v'(x)$

   Here, $u(x) = x^2$ and $v(x) = (1 - 4x)^4$.

2. Differentiate $u(x) = x^2$:

   $u'(x) = 2x$

3. Differentiate $v(x) = (1 - 4x)^4$: We use the chain rule for this part. If $v(x) = (f(x))^4$ where $f(x) = 1 - 4x$, the chain rule gives us:

   $v'(x) = 4(1 - 4x)^3 \cdot (-4)$

   So,

   $v'(x) = -16(1 - 4x)^3$

4. Apply the product rule:

   $\frac{dy}{dx} = u'(x) v(x) + u(x) v'(x)$

   Substituting the values for $u'(x)$, $v(x)$, and $v'(x)$:

   $\frac{dy}{dx} = 2x(1 - 4x)^4 + x^2 \cdot (-16)(1 - 4x)^3$

So, the derivative is:

$\frac{dy}{dx} = 2x(1 - 4x)^4 - 16x^2(1 - 4x)^3$

Let me know if you'd like further explanations or clarifications!

Related Questions:

1. How do you apply the chain rule in differentiation?
2. What is the product rule, and can it be used for more than two functions?
3. How do you factor expressions like $(1 - 4x)^4 - 16x^2(1 - 4x)^3$?
4. How do you simplify derivatives when there's a common factor?
5. What is the chain rule used for in other areas of calculus, like integration?

Tip:

When using the chain rule, remember to differentiate the outer function first, then multiply by the derivative of the inner function.