Let's simplify the expression:

$f'(x) = \frac{2e^{2x} \cdot (8e^{2x} - 2e^x) - (4e^{2x} - 2e^x + 1) \cdot 4e^{2x}}{(2e^{2x})^2}$

Step-by-Step Simplification:

1. Simplify the denominator: $(2e^{2x})^2 = 4e^{4x}$ So the expression becomes: $f'(x) = \frac{2e^{2x} \cdot (8e^{2x} - 2e^x) - (4e^{2x} - 2e^x + 1) \cdot 4e^{2x}}{4e^{4x}}$

2. Expand the terms in the numerator:

   • For the first term: = 16e^{4x} - 4e^{3x}$$
   • For the second term: = 16e^{4x} - 8e^{3x} + 4e^{2x}$$

3. Combine these two results: The numerator is now: $(16e^{4x} - 4e^{3x}) - (16e^{4x} - 8e^{3x} + 4e^{2x})$ Simplifying this: $= 16e^{4x} - 4e^{3x} - 16e^{4x} + 8e^{3x} - 4e^{2x}$ Combine like terms:

   = 4e^{3x} - 4e^{2x}$$

4. Rewrite the entire expression: $f'(x) = \frac{4e^{3x} - 4e^{2x}}{4e^{4x}}$

5. Simplify the fraction: Factor out the common term $4e^{2x}$ in the numerator: $f'(x) = \frac{4e^{2x}(e^x - 1)}{4e^{4x}}$ Cancel out the $4e^{2x}$ terms: $f'(x) = \frac{e^x - 1}{e^{2x}}$

6. Final simplified expression: $f'(x) = \frac{e^x - 1}{e^{2x}} = e^{-x} - e^{-2x}$

The simplified expression for $f'(x)$ is:

$f'(x) = e^{-x} - e^{-2x}$

Would you like more details on any of the steps, or have any questions?

Here are some related questions for further exploration:

1. How would you apply the chain rule to differentiate exponential functions?
2. Can you verify the simplified derivative by differentiating the original function?
3. How does the behavior of $e^{-x} - e^{-2x}$ change as $x \to \infty$?
4. What is the integral of $f'(x) = e^{-x} - e^{-2x}$?
5. How would you plot the function and its derivative to compare their behaviors?

Tip: When simplifying complex expressions, look for common factors in the numerator and denominator to make the process more efficient.