We are given:

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

To find $\frac{dy}{dx}$, we will use the chain rule and differentiate the function step by step.

Step 1: Let’s start by differentiating the outer function

The outer function is $y = \tan(u)$, where $u = 2 \tan^{-1}\left(\frac{x}{2}\right)$.

The derivative of $\tan(u)$ is:

$\frac{dy}{du} = \sec^2(u)$

Step 2: Differentiate the inner function $u = 2 \tan^{-1}\left(\frac{x}{2}\right)$

We need to find $\frac{du}{dx}$.

First, recall that the derivative of $\tan^{-1}(z)$ with respect to $z$ is:

$\frac{d}{dz} \left( \tan^{-1}(z) \right) = \frac{1}{1 + z^2}$

Now apply the chain rule. If $z = \frac{x}{2}$, then:

$\frac{d}{dx} \left( \tan^{-1} \left( \frac{x}{2} \right) \right) = \frac{1}{1 + \left( \frac{x}{2} \right)^2} \cdot \frac{1}{2}$

Simplifying this:

$\frac{d}{dx} \left( \tan^{-1} \left( \frac{x}{2} \right) \right) = \frac{1}{2 \left(1 + \frac{x^2}{4} \right)} = \frac{2}{4 + x^2}$

Now differentiate $u = 2 \tan^{-1} \left( \frac{x}{2} \right)$:

$\frac{du}{dx} = 2 \cdot \frac{2}{4 + x^2} = \frac{4}{4 + x^2}$

Step 3: Apply the chain rule

Now that we have $\frac{dy}{du} = \sec^2(u)$ and $\frac{du}{dx} = \frac{4}{4 + x^2}$, we can apply the chain rule:

$\frac{dy}{dx} = \sec^2(u) \cdot \frac{du}{dx}$

Substitute $u = 2 \tan^{-1} \left( \frac{x}{2} \right)$ back into the equation:

$\frac{dy}{dx} = \sec^2\left( 2 \tan^{-1}\left( \frac{x}{2} \right) \right) \cdot \frac{4}{4 + x^2}$

This is the final derivative.

Would you like more details or a clarification on any step?

Here are five related questions for further exploration:

1. How is the chain rule applied in the differentiation of composite functions?
2. What is the derivative of the inverse tangent function, and how is it derived?
3. Can you apply the same technique to differentiate functions involving $\sin^{-1}(x)$ or $\cos^{-1}(x)$?
4. How does the identity $\sec^2(x) = 1 + \tan^2(x)$ help in solving trigonometric derivatives?
5. What are the applications of the inverse trigonometric functions in calculus?

Tip: Always break down complex functions into simpler parts to apply the chain rule effectively.