Let's carefully analyze the steps shown in the expression. The expression is:

$y = \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} = \frac{e^{2x} - \frac{1}{e^{2x}}}{e^{2x} + \frac{1}{e^{2x}}}$

We want to see how $e^{2x}$ converts into $e^{4x}$.

1. In the third step, we multiply both the numerator and denominator of the fraction by $e^{2x}$ to eliminate the terms with $e^{-2x}$. This is a common algebraic technique to simplify the expression.

Step-by-Step Simplification:

Starting expression:

$\frac{e^{2x} - \frac{1}{e^{2x}}}{e^{2x} + \frac{1}{e^{2x}}}$

Multiply both numerator and denominator by $e^{2x}$:

• Numerator: $e^{2x} \left( e^{2x} - \frac{1}{e^{2x}} \right) = e^{4x} - 1$

• Denominator: $e^{2x} \left( e^{2x} + \frac{1}{e^{2x}} \right) = e^{4x} + 1$

Resulting expression:

$y = \frac{e^{4x} - 1}{e^{4x} + 1}$

Thus, the term $e^{2x}$ transforms into $e^{4x}$ by multiplying both the numerator and the denominator by $e^{2x}$.

Would you like more details on this simplification, or do you have any other questions?

Here are 5 related questions to deepen understanding:

1. How does multiplying both numerator and denominator by the same term affect a fraction?
2. What are the rules of exponents used in this simplification?
3. How would the expression change if we multiplied by a different power of $e$?
4. Can we simplify this further by introducing hyperbolic functions like $\tanh(x)$?
5. What happens to this function as $x \to \infty$?

Tip: Multiplying both the numerator and denominator by a common factor is a useful technique when trying to eliminate negative exponents or simplify complex fractions.