We are tasked with finding the derivative of the function:

$y = \left( \tan^{-1}(4x) \right)^2$

Step-by-step solution:

1. Apply the chain rule: To differentiate this expression, we treat the outer function as $u^2$ where $u = \tan^{-1}(4x)$. So the derivative of $y = u^2$ with respect to $u$ is: $\frac{d}{dx}[u^2] = 2u \cdot \frac{du}{dx}$ Therefore: $y' = 2 \cdot \tan^{-1}(4x) \cdot \frac{d}{dx}\left[ \tan^{-1}(4x) \right]$

2. Differentiate $\tan^{-1}(4x)$: Use the chain rule again to differentiate $\tan^{-1}(4x)$. The derivative of $\tan^{-1}(u)$ is $\frac{1}{1 + u^2}$, and since $u = 4x$, we also multiply by the derivative of $4x$ (which is 4): $\frac{d}{dx}\left[ \tan^{-1}(4x) \right] = \frac{4}{1 + (4x)^2} = \frac{4}{1 + 16x^2}$

3. Combine the results: Now substitute this back into the expression for $y'$: $y' = 2 \cdot \tan^{-1}(4x) \cdot \frac{4}{1 + 16x^2}$ Simplifying: $y' = \frac{8 \cdot \tan^{-1}(4x)}{1 + 16x^2}$

Thus, the derivative of the given function is:

$y' = \frac{8 \cdot \tan^{-1}(4x)}{1 + 16x^2}$

Would you like more details or clarification on any step?

Here are five related questions to expand your understanding:

1. What are the properties of the inverse tangent function, $\tan^{-1}(x)$?
2. How does the chain rule work for compositions of functions?
3. Can you explain why the derivative of $\tan^{-1}(x)$ is $\frac{1}{1 + x^2}$?
4. How would the derivative change if the argument inside the inverse tangent were a more complex function of $x$?
5. Can we apply a similar method to differentiate powers of other inverse trigonometric functions?

Tip: When differentiating functions that involve inverse trigonometric expressions, always pay close attention to the chain rule and any inner functions you need to differentiate.