We are asked to find the derivative \( \frac{dy}{dx} \) for the equation: 

\[
\tan(x - y) = y
\]

Since \( y \) is implicitly defined in terms of \( x \), we will use **implicit differentiation**.

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### Steps for Implicit Differentiation:

1. Differentiate both sides of the equation with respect to \( x \).
   - Use the **chain rule** for \( \tan(x - y) \) and remember that \( y \) is a function of \( x \).

\[
\frac{d}{dx}[\tan(x - y)] = \frac{d}{dx}[y]
\]

2. Apply the derivative rules:
   - \( \frac{d}{dx}[\tan(u)] = \sec^2(u) \cdot \frac{du}{dx} \).
   - For the right side, \( \frac{d}{dx}[y] = \frac{dy}{dx} \).

\[
\sec^2(x - y) \cdot \left(\frac{d}{dx}(x - y)\right) = \frac{dy}{dx}
\]

3. Differentiate \( x - y \) with respect to \( x \):
   - \( \frac{d}{dx}(x - y) = 1 - \frac{dy}{dx} \).

\[
\sec^2(x - y) \cdot (1 - \frac{dy}{dx}) = \frac{dy}{dx}
\]

4. Expand the equation:

\[
\sec^2(x - y) - \sec^2(x - y) \cdot \frac{dy}{dx} = \frac{dy}{dx}
\]

5. Collect all \( \frac{dy}{dx} \)-terms on one side:

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

6. Factor out \( \frac{dy}{dx} \) on the right side:

\[
\sec^2(x - y) = \frac{dy}{dx} \left(1 + \sec^2(x - y)\right)
\]

7. Solve for \( \frac{dy}{dx} \):

\[
\frac{dy}{dx} = \frac{\sec^2(x - y)}{1 + \sec^2(x - y)}
\]

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### Final Answer:

\[
\frac{dy}{dx} = \frac{\sec^2(x - y)}{1 + \sec^2(x - y)}
\]

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Would you like further clarification or examples? 

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Here are 5 related questions:
1. How do you apply implicit differentiation for other trigonometric functions?
2. Can we solve for \( y \) explicitly in this problem?
3. What is the significance of the chain rule in implicit differentiation?
4. How do we interpret the behavior of \( \frac{dy}{dx} \) at different points?
5. How does the derivative change if the equation involves \( \cot(x - y) \) instead?

**Tip:** When applying implicit differentiation, carefully apply the chain rule whenever a function of \( y \) is involved.

Explore how to find the derivative dy/dx using implicit differentiation for the equation tan(x - y) = y. This step-by-step guide covers key concepts like the chain rule and trigonometric derivatives, suitable for students in Grades 11-12 or Calculus I.

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