You want to find the \(n\)th derivative of the function \(x^{2n}\) with respect to \(x\). Let's break this down:

Given the function \(f(x) = x^{2n}\), we want to find \(\frac{d^n}{dx^n} x^{2n}\).

### Step-by-Step Solution:

1. **First Derivative:**
   \[
   \frac{d}{dx} x^{2n} = 2n \cdot x^{2n-1}
   \]

2. **Second Derivative:**
   \[
   \frac{d^2}{dx^2} x^{2n} = 2n \cdot (2n-1) \cdot x^{2n-2}
   \]

3. **Third Derivative:**
   \[
   \frac{d^3}{dx^3} x^{2n} = 2n \cdot (2n-1) \cdot (2n-2) \cdot x^{2n-3}
   \]

4. **General \(n\)th Derivative:**
   The pattern here is that each derivative introduces a factor of \(2n - (k-1)\), where \(k\) is the number of derivatives taken so far. For the \(n\)th derivative, the general formula becomes:

   \[
   \frac{d^n}{dx^n} x^{2n} = 2n \cdot (2n-1) \cdot (2n-2) \cdot \ldots \cdot (2n-n+1) \cdot x^{2n-n}
   \]

   Simplifying this, we get:
   \[
   \frac{d^n}{dx^n} x^{2n} = \frac{(2n)!}{(2n-n)!} \cdot x^n = \frac{(2n)!}{n!} \cdot x^n
   \]

### Final Answer:
\[
\frac{d^n}{dx^n} x^{2n} = \frac{(2n)!}{n!} \cdot x^n
\]

This is the \(n\)th derivative of \(x^{2n}\) with respect to \(x\).

Would you like more details or have any questions about this process? Here are five related questions you might consider:

1. How would this approach change for finding higher derivatives of other powers of \(x\)?
2. What happens if you take the \(n\)th derivative of a function like \(x^m\) where \(m \neq 2n\)?
3. Can you derive the expression for \(\frac{d^n}{dx^n} x^{m}\) in general?
4. What are the applications of taking higher derivatives in calculus?
5. How does the factorial in the result relate to combinatorics?

**Tip:** When calculating higher-order derivatives, look for patterns in the coefficients that can help simplify the general formula.

Explore how to calculate the nth derivative of the function x^(2n) with a detailed step-by-step solution. This problem covers key calculus concepts, including derivatives and factorials. Suitable for advanced undergraduate students and beyond.

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