To differentiate the function \( e^{2-x} \cdot \tan^2{x} \), we will apply both the product rule and the chain rule.

### Given:
\[
f(x) = e^{2-x} \cdot \tan^2{x}
\]

### Steps:
1. **Product rule**: For two functions \( u(x) \) and \( v(x) \), the product rule is given by:
   \[
   \frac{d}{dx}(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)
   \]
   Here, \( u(x) = e^{2-x} \) and \( v(x) = \tan^2{x} \).

2. **Differentiate \( u(x) = e^{2-x} \)**:
   \[
   u'(x) = \frac{d}{dx} \left( e^{2-x} \right) = e^{2-x} \cdot (-1) = -e^{2-x}
   \]
   
3. **Differentiate \( v(x) = \tan^2{x} \)**:
   Using the chain rule, we get:
   \[
   v'(x) = 2\tan{x} \cdot \sec^2{x}
   \]

4. **Apply the product rule**:
   \[
   \frac{d}{dx} \left( e^{2-x} \cdot \tan^2{x} \right) = u'(x) \cdot v(x) + u(x) \cdot v'(x)
   \]
   Substituting the derivatives:
   \[
   \frac{d}{dx} \left( e^{2-x} \cdot \tan^2{x} \right) = \left( -e^{2-x} \cdot \tan^2{x} \right) + \left( e^{2-x} \cdot 2\tan{x} \cdot \sec^2{x} \right)
   \]

### Final result:
\[
\frac{d}{dx} \left( e^{2-x} \cdot \tan^2{x} \right) = -e^{2-x} \cdot \tan^2{x} + 2e^{2-x} \cdot \tan{x} \cdot \sec^2{x}
\]

Let me know if you want further clarification or a more detailed breakdown of any step.

### Related Questions:
1. How do we apply the product rule to more than two terms?
2. What is the chain rule, and when is it used?
3. Can you explain the derivative of \( \tan{x} \) and \( \sec{x} \) in detail?
4. How do exponential functions behave under differentiation?
5. Why do we use the product rule in differentiation?

**Tip**: Always break complex derivatives into smaller parts and apply rules step by step to avoid errors.

Learn how to differentiate the function e^(2-x) * tan^2(x) using the product and chain rule. This calculus problem provides a step-by-step breakdown and includes the use of exponential and trigonometric derivatives. Suitable for students in Grades 11-12.

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