The problem asks us to find the \( n \)th derivative of the function:

\[
f(x) = \frac{1}{1 - 5x + 6x^2}.
\]

Let's break down the steps:

### 1. Factor the Denominator

First, note that the denominator is a quadratic expression. Let's factor it:

\[
1 - 5x + 6x^2 = (1 - 2x)(1 - 3x).
\]

So, the function can be rewritten as:

\[
f(x) = \frac{1}{(1 - 2x)(1 - 3x)}.
\]

### 2. Express as a Partial Fraction

We can express the function as a sum of partial fractions:

\[
f(x) = \frac{A}{1 - 2x} + \frac{B}{1 - 3x},
\]

where \( A \) and \( B \) are constants to be determined. To find \( A \) and \( B \), we equate:

\[
\frac{1}{(1 - 2x)(1 - 3x)} = \frac{A}{1 - 2x} + \frac{B}{1 - 3x}.
\]

Multiplying through by the denominator gives:

\[
1 = A(1 - 3x) + B(1 - 2x).
\]

Expanding this expression:

\[
1 = A - 3Ax + B - 2Bx.
\]

Grouping terms gives:

\[
1 = (A + B) + (-3A - 2B)x.
\]

For this to hold for all \( x \), the coefficients of \( x \) and the constant term must be equal:

1. \( A + B = 1 \).
2. \( -3A - 2B = 0 \).

Solving these equations simultaneously:

From \( -3A - 2B = 0 \), we get \( B = -\frac{3}{2}A \).

Substituting into \( A + B = 1 \):

\[
A - \frac{3}{2}A = 1 \implies -\frac{1}{2}A = 1 \implies A = -2.
\]

Then, \( B = -\frac{3}{2}(-2) = 3 \).

So, the partial fraction decomposition is:

\[
f(x) = \frac{-2}{1 - 2x} + \frac{3}{1 - 3x}.
\]

### 3. Express as a Series

Each term can be expressed as a geometric series:

\[
\frac{1}{1 - ax} = \sum_{k=0}^{\infty} (ax)^k.
\]

Applying this to our expression:

\[
f(x) = -2 \sum_{k=0}^{\infty} (2x)^k + 3 \sum_{k=0}^{\infty} (3x)^k.
\]

Simplifying:

\[
f(x) = \sum_{k=0}^{\infty} \left[ -2 \cdot 2^k + 3 \cdot 3^k \right] x^k.
\]

### 4. Differentiate to Find the \( n \)th Derivative

The \( n \)th derivative of \( x^k \) is given by:

\[
\frac{d^n}{dx^n} x^k = \begin{cases} 
\frac{k!}{(k-n)!} x^{k-n} & \text{if } k \geq n, \\
0 & \text{if } k < n.
\end{cases}
\]

So, the \( n \)th derivative of \( f(x) \) is:

\[
f^{(n)}(x) = \sum_{k=n}^{\infty} \left[ -2 \cdot 2^k + 3 \cdot 3^k \right] \frac{k!}{(k-n)!} x^{k-n}.
\]

Evaluating this at \( x = 0 \) gives:

\[
f^{(n)}(0) = \sum_{k=n}^{\infty} \left[ -2 \cdot 2^k + 3 \cdot 3^k \right] \frac{k!}{(k-n)!}.
\]

This expression gives the \( n \)th derivative of the function at \( x = 0 \).

Would you like more details or have any other questions?

Here are some related questions:
1. How do you compute the second derivative of a function?
2. What is the radius of convergence for a geometric series?
3. How do you find the partial fraction decomposition of a function?
4. How do you compute the series expansion for a function?
5. What are the applications of geometric series in calculus?

**Tip:** In partial fractions, always check for factors in the denominator that can simplify your problem before expanding.

Learn how to find the nth derivative of the function 1/(1 - 5x + 6x^2) using differentiation, partial fraction decomposition, and series expansion. This problem covers key calculus concepts, including geometric series, power function derivatives, and the binomial theorem. Suitable for advanced undergraduate students.

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