Given the function $y = 2x e^x$, you are asked to find $y^n$, which typically refers to the $n$-th derivative of $y$ with respect to $x$.

Steps to Calculate the $n$-th Derivative

1. First Derivative $y'$ (or $y^{(1)}$): $y' = \frac{d}{dx}(2x e^x)$ We apply the product rule: $y' = 2 \left( \frac{d}{dx}(x) \cdot e^x + x \cdot \frac{d}{dx}(e^x) \right) = 2(e^x + x e^x) = 2e^x(1 + x)$

2. Second Derivative $y''$ (or $y^{(2)}$): $y'' = \frac{d}{dx}\left( 2e^x(1 + x) \right)$ Again, using the product rule: $y'' = 2 \left( \frac{d}{dx}(e^x) \cdot (1 + x) + e^x \cdot \frac{d}{dx}(1 + x) \right) = 2 \left( e^x(1 + x) + e^x \right) = 2e^x(2 + x)$

3. Third Derivative $y'''$ (or $y^{(3)}$): $y''' = \frac{d}{dx}\left( 2e^x(2 + x) \right)$ Again, using the product rule: $y''' = 2 \left( e^x(2 + x) + e^x \right) = 2e^x(3 + x)$

General Pattern for the $n$-th Derivative:

Observing the pattern from the first few derivatives, the general form of the $n$-th derivative appears to be: $y^{(n)} = 2e^x(n + x)$ This pattern holds because with each differentiation, the coefficient in front of $e^x$ increases by 1.

Final Answer:

$y^{(n)} = 2e^x(n + x)$

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. What happens if the exponent of $e^x$ in the original function is changed?
2. How would the derivative process differ if there was an additional trigonometric factor?
3. Can you find the general derivative of a function of the form $f(x) = x^n e^x$?
4. What role does the product rule play in finding derivatives?
5. How does differentiation affect the exponential function?

Tip: When finding higher-order derivatives, look for patterns in the coefficients and terms to simplify the process.