If f(x) = log_{(x^2-2x+1)}(x^2-3x+2),x in R - | vedclass

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(D) Given $f(x) = \log_{(x^2-2x+1)}(x^2-3x+2)$.We can rewrite the base and the argument as:$x^2-2x+1 = (x-1)^2$$x^2-3x+2 = (x-1)(x-2)$So,$f(x) = \log_

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$\log_4 e$

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(D) Given $f(x) = \log_{(x^2-2x+1)}(x^2-3x+2)$.We can rewrite the base and the argument as:$x^2-2x+1 = (x-1)^2$$x^2-3x+2 = (x-1)(x-2)$So,$f(x) = \log_{(x-1)^2} ((x-1)(x-2))$.Using the property $\log_{a^n} b = \frac{1}{n} \log_a b$,we get:$f(x) = \frac{1}{2} \log_{(x-1)} ((x-1)(x-2)) = \frac{1}{2} [\log_{(x-1)} (x-1) + \log_{(x-1)} (x-2)]$$f(x) = \frac{1}{2} [1 + \log_{(x-1)} (x-2)] = \frac{1}{2} [1 + \frac{\ln(x-2)}{\ln(x-1)}]$.Now,differentiate with respect to $x$:$f'(x) = \frac{1}{2} \left[ \frac{\ln(x-1) \cdot \frac{1}{x-2} - \ln(x-2) \cdot \frac{1}{x-1}}{(\ln(x-1))^2} \right]$.Substitute $x = 3$:$f'(3) = \frac{1}{2} \left[ \frac{\ln(2) \cdot \frac{1}{1} - \ln(1) \cdot \frac{1}{2}}{(\ln(2))^2} \right] = \frac{1}{2} \left[ \frac{\ln(2)}{(\ln(2))^2} \right] = \frac{1}{2 \ln(2)} = \frac{1}{\ln(4)} = \log_4 e$.

If $f(x) = \log_{(x^2-2x+1)}(x^2-3x+2)$,$x \in R - \{1, 2\}$ and $x \neq 0$,then $f'(3) =$

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