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{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.