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To solve the problem given in the image, we need to find the 101st derivative of the function \( f(x) \) at \( x = 0 \). The function is defined as:

\[ f(x) = \frac{x}{1 - 2x^4} \]

First, let's rewrite the function in a more convenient form for differentiation. We can use a power series expansion:

\[ f(x) = x \cdot \frac{1}{1 - 2x^4} \]

We recognize the geometric series sum:

\[ \frac{1}{1 - 2x^4} = \sum_{n=0}^{\infty} (2x^4)^n = \sum_{n=0}^{\infty} 2^n x^{4n} \]

Therefore,

\[ f(x) = x \cdot \sum_{n=0}^{\infty} 2^n x^{4n} = \sum_{n=0}^{\infty} 2^n x^{4n+1} \]

We need to find the coefficient of \( x^{101} \) in this series because that coefficient will be the 101st derivative of \( f(x) \) evaluated at \( x = 0 \) divided by 101 factorial.

In the series \( \sum_{n=0}^{\infty} 2^n x^{4n+1} \), the term with \( x^{101} \) occurs when:

\[ 4n + 1 = 101 \]

Solving for \( n \):

\[ 4n + 1 = 101 \]
\[ 4n = 100 \]
\[ n = 25 \]

So the coefficient of \( x^{101} \) is \( 2^{25} \). Therefore, the 101st derivative of \( f(x) \) at \( x = 0 \) is:

\[ f^{(101)}(0) = 101! \cdot 2^{25} \]

Thus,

\[ f^{(101)}(0) = 2^{25} \cdot 101! \]

Would you like more details or have any other questions? Here are some follow-up questions you might consider:

1. How do you find the radius of convergence for the given power series?
2. What are some applications of higher-order derivatives in real-world problems?
3. Can you explain how to derive the Taylor series expansion for different types of functions?
4. What methods can be used to simplify the calculation of high-order derivatives?
5. How does the presence of singularities affect the power series representation of a function?

**Tip:** When working with power series, always look for opportunities to simplify the series using known expansions, such as the geometric series, which can save significant time and effort.

Learn how to find the 101st derivative of the function f(x) = x / (1 - 2x^4) using power series expansion and geometric series. This problem involves advanced calculus concepts and is suitable for undergraduate students.

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