To find the derivative of the equation $\tan(2x + y) = 2x$, we will use implicit differentiation. Here’s the process step-by-step:

1. Differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$. Use the chain rule on the left-hand side, because $y$ depends on $x$, and the derivative of $\tan(u)$ is $\sec^2(u) \cdot u'$.

$\frac{d}{dx} \left( \tan(2x + y) \right) = \frac{d}{dx}(2x)$

2. Applying the chain rule to the left-hand side: $\sec^2(2x + y) \cdot \frac{d}{dx}(2x + y) = 2$

3. Now, differentiate $2x + y$ with respect to $x$: $\frac{d}{dx}(2x + y) = 2 + \frac{dy}{dx}$

4. Substitute this into the equation: $\sec^2(2x + y) \cdot (2 + \frac{dy}{dx}) = 2$

5. Now, solve for $\frac{dy}{dx}$:

$\sec^2(2x + y) \cdot (2 + \frac{dy}{dx}) = 2$

Divide both sides by $\sec^2(2x + y)$:

$2 + \frac{dy}{dx} = \frac{2}{\sec^2(2x + y)} = 2 \cos^2(2x + y)$

6. Isolate $\frac{dy}{dx}$:

$\frac{dy}{dx} = 2 \cos^2(2x + y) - 2$

Therefore, the derivative is: $\frac{dy}{dx} = 2(\cos^2(2x + y) - 1)$

Do you want details on the steps, or any specific clarifications?

Here are 5 questions to expand on this:

1. What is the chain rule in differentiation?
2. How does implicit differentiation differ from explicit differentiation?
3. Why do we treat $y$ as a function of $x$ in this context?
4. How does the derivative of $\tan(u)$ relate to $\sec^2(u)$?
5. Can you derive the identity $2 \cos^2(x) - 1$?

Tip: When differentiating trigonometric functions, keep the derivatives of basic trig functions like $\sin(x)$, $\cos(x)$, and $\tan(x)$ at hand for quicker reference.