Math Problem Statement

d^n/dx^n x^2n

Solution

You want to find the nnth derivative of the function x2nx^{2n} with respect to xx. Let's break this down:

Given the function f(x)=x2nf(x) = x^{2n}, we want to find dndxnx2n\frac{d^n}{dx^n} x^{2n}.

Step-by-Step Solution:

  1. First Derivative: ddxx2n=2nx2n1\frac{d}{dx} x^{2n} = 2n \cdot x^{2n-1}

  2. Second Derivative: d2dx2x2n=2n(2n1)x2n2\frac{d^2}{dx^2} x^{2n} = 2n \cdot (2n-1) \cdot x^{2n-2}

  3. Third Derivative: d3dx3x2n=2n(2n1)(2n2)x2n3\frac{d^3}{dx^3} x^{2n} = 2n \cdot (2n-1) \cdot (2n-2) \cdot x^{2n-3}

  4. General nnth Derivative: The pattern here is that each derivative introduces a factor of 2n(k1)2n - (k-1), where kk is the number of derivatives taken so far. For the nnth derivative, the general formula becomes:

    dndxnx2n=2n(2n1)(2n2)(2nn+1)x2nn\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: dndxnx2n=(2n)!(2nn)!xn=(2n)!n!xn\frac{d^n}{dx^n} x^{2n} = \frac{(2n)!}{(2n-n)!} \cdot x^n = \frac{(2n)!}{n!} \cdot x^n

Final Answer:

dndxnx2n=(2n)!n!xn\frac{d^n}{dx^n} x^{2n} = \frac{(2n)!}{n!} \cdot x^n

This is the nnth derivative of x2nx^{2n} with respect to xx.

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 xx?
  2. What happens if you take the nnth derivative of a function like xmx^m where m2nm \neq 2n?
  3. Can you derive the expression for dndxnxm\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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Factorials

Formulas

General formula for the nth derivative

Theorems

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Suitable Grade Level

Advanced Undergraduate