Step 1: Simplify the function
The base of the logarithm is \( x^2-2x+1 = (x-1)^2 \).
The argument is \( x^2-3x+2 = (x-1)(x-2) \).
Using the change of base property \( \log_a b = \frac{\ln b}{\ln a} \):
\[ f(x) = \frac{\ln((x-1)(x-2))}{\ln((x-1)^2)} = \frac{\ln(x-1) + \ln(x-2)}{2\ln(x-1)} \]
\[ f(x) = \frac{1}{2} + \frac{1}{2} \frac{\ln(x-2)}{\ln(x-1)} \]
Step 2: Differentiate the function
Use the quotient rule for the second term:
\[ f'(x) = \frac{1}{2} \left[ \frac{\ln(x-1) \cdot \frac{1}{x-2} - \ln(x-2) \cdot \frac{1}{x-1}}{(\ln(x-1))^2} \right] \]
Step 3: Calculate \( f'(3) \)
Substitute \( x = 3 \):
Terms involved:
\( x-1 = 2 \), \( \ln(x-1) = \ln 2 \).
\( x-2 = 1 \), \( \ln(x-2) = \ln 1 = 0 \).
Substitute these values:
\[ f'(3) = \frac{1}{2} \left[ \frac{(\ln 2) \cdot \frac{1}{1} - 0 \cdot \frac{1}{2}}{(\ln 2)^2} \right] \]
\[ f'(3) = \frac{1}{2} \frac{\ln 2}{(\ln 2)^2} = \frac{1}{2 \ln 2} = \frac{1}{\ln 4} \]
Since \( \frac{1}{\ln 4} = \log_4 e \):
\[ f'(3) = \log_4 e \]