If` f(x) = (x - 1)/(4) + ((x - 1)^(2))/(12) + ((x -1)^(5))/(20) + ((x -1)^(7))/(28) + ... `
where ` 0 lt x lt 2 ` , then f'(x) is equal to

Options

A. `(1)/(4x(2-x)) `
B. `(1)/(4(x-2)^(2))`
C. `(1)/(2-x)`
D. `(1)/(2+x)`

Explanation

To find the derivative \( f'(x) \) of the function

\[ f(x) = \frac{x - 1}{4} + \frac{(x - 1)^2}{12} + \frac{(x - 1)^5}{20} + \frac{(x - 1)^7}{28} + \ldots \]

we can observe that this series can be expressed in a more manageable form.

### Step 1: Recognize the Series
The series can be rewritten in terms of a general term. The general term appears to be:

\[ \frac{(x - 1)^{n}}{a_n} \]

where \( n \) takes on odd values (1, 2, 5, 7, ...), and \( a_n \) is a sequence of coefficients (4, 12, 20, 28, ...).

### Step 2: Identify the Pattern
The coefficients can be recognized as:

- \( a_1 = 4 = 2 \times 2 \)
- \( a_2 = 12 = 3 \times 4 \)
- \( a_3 = 20 = 4 \times 5 \)
- \( a_4 = 28 = 4 \times 7 \)

This suggests that the series might be related to a logarithmic function.

### Step 3: Relate to Logarithmic Expansion
The series can be recognized as a Taylor series expansion of a logarithmic function. Specifically, the series resembles the expansion of:

\[ \log\left(\frac{1+t}{1-t}\right) = 2\left(t + \frac{t^3}{3} + \frac{t^5}{5} + \ldots\right) \]

where \( t = x - 1 \).

### Step 4: Express \( f(x) \)
Thus, we can express \( f(x) \) as:

\[ f(x) = \frac{1}{8} \log\left(\frac{1 + (x-1)}{1 - (x-1)}\right) = \frac{1}{8} \log\left(\frac{x}{2 - x}\right) \]

### Step 5: Differentiate \( f(x) \)
Now, we differentiate \( f(x) \):

\[ f'(x) = \frac{1}{8} \cdot \frac{d}{dx} \left( \log\left(\frac{x}{2 - x}\right) \right) \]

Using the chain rule, we have:

\[ f'(x) = \frac{1}{8} \cdot \frac{1}{\frac{x}{2-x}} \cdot \frac{d}{dx}\left(\frac{x}{2 - x}\right) \]

### Step 6: Differentiate the Inner Function
To differentiate \( \frac{x}{2-x} \), we use the quotient rule:

\[ \frac{d}{dx}\left(\frac{x}{2 - x}\right) = \frac{(2-x)(1) - x(-1)}{(2-x)^2} = \frac{2 - x + x}{(2-x)^2} = \frac{2}{(2-x)^2} \]

### Step 7: Combine the Results
Now substituting back, we get:

\[ f'(x) = \frac{1}{8} \cdot \frac{2}{\frac{x}{2-x}} \cdot \frac{2}{(2-x)^2} = \frac{1}{4} \cdot \frac{2(2-x)}{x(2-x)^2} \]

### Final Result
Thus, simplifying gives us:

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

### Summary
The derivative \( f'(x) \) is:

\[ f'(x) = \frac{2 - x}{2x(2 - x)} \]