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.

Discover how to apply implicit differentiation to solve the equation tan(2x + y) = 2x. This detailed solution covers calculus concepts like the chain rule and trigonometric derivatives. Perfect for students in Grades 11-12 or introductory college courses.

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