Question

If $f(x) = \log_{(x-1)^2}(x^2-3x+2)$, $x \in \mathbb{R}-[1,2]$ and $x\neq0$, then $f'(3)=$

• A. 1
• B. 0
• C. $\log_e 4$
• D. $\log_4 e$

Hint:

When dealing with logarithmic functions with a variable in the base, the first step should always be to use the change of base formula to convert to a standard base like $e$ or 10. This makes differentiation much more straightforward.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The problem requires finding the derivative of a function involving a logarithm with a variable base and then evaluating it at a specific point. The first step is to simplify the function using logarithm properties, particularly the change of base formula. 
Step 2: Key Formula or Approach 
1. Use the change of base formula for logarithms: $\log_b a = \frac{\ln a}{\ln b}$. 2. Simplify the resulting expression using properties of logarithms like $\ln(ab) = \ln a + \ln b$ and $\ln(a^n) = n\ln a$. 3. Differentiate the simplified function using the quotient rule: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$. 4. Substitute $x=3$ into the derivative to find the final value. 
Step 3: Detailed Explanation 
1. Simplify the function $f(x)$: The function is $f(x) = \log_{(x-1)^2}(x^2-3x+2)$. First, factor the argument of the logarithm: $x^2-3x+2 = (x-1)(x-2)$. \[ f(x) = \log_{(x-1)^2}((x-1)(x-2)) \] Apply the change of base formula to natural logarithm (ln): \[ f(x) = \frac{\ln((x-1)(x-2))}{\ln((x-1)^2)} \] Using logarithm properties: \[ f(x) = \frac{\ln(x-1) + \ln(x-2)}{2\ln(x-1)} \] (The domain $x \in \mathbb{R}-[1,2]$ ensures that the arguments of the logarithms are well-defined and positive for $x>2$). Split the fraction: \[ f(x) = \frac{\ln(x-1)}{2\ln(x-1)} + \frac{\ln(x-2)}{2\ln(x-1)} = \frac{1}{2} + \frac{1}{2}\frac{\ln(x-2)}{\ln(x-1)} \] 
2. Differentiate $f(x)$: \[ f'(x) = \frac{d}{dx}\left(\frac{1}{2} + \frac{1}{2}\frac{\ln(x-2)}{\ln(x-1)}\right) = 0 + \frac{1}{2} \frac{d}{dx}\left(\frac{\ln(x-2)}{\ln(x-1)}\right) \] Using the quotient rule for the fraction: \[ f'(x) = \frac{1}{2} \left[ \frac{\frac{d}{dx}(\ln(x-2)) \cdot \ln(x-1) - \ln(x-2) \cdot \frac{d}{dx}(\ln(x-1))}{(\ln(x-1))^2} \right] \] \[ f'(x) = \frac{1}{2} \left[ \frac{\frac{1}{x-2} \cdot \ln(x-1) - \ln(x-2) \cdot \frac{1}{x-1}}{(\ln(x-1))^2} \right] \] 
3. Evaluate $f'(3)$: Substitute $x=3$ into the expression for $f'(x)$: \[ f'(3) = \frac{1}{2} \left[ \frac{\frac{1}{3-2} \cdot \ln(3-1) - \ln(3-2) \cdot \frac{1}{3-1}}{(\ln(3-1))^2} \right] \] \[ f'(3) = \frac{1}{2} \left[ \frac{1 \cdot \ln(2) - \ln(1) \cdot \frac{1}{2}}{(\ln 2)^2} \right] \] Since $\ln(1)=0$: \[ f'(3) = \frac{1}{2} \left[ \frac{\ln 2 - 0}{(\ln 2)^2} \right] = \frac{1}{2} \left[ \frac{\ln 2}{(\ln 2)^2} \right] = \frac{1}{2\ln 2} \] 
4. Match with options: The result is $\frac{1}{2\ln 2}$. We can rewrite this using logarithm properties: \[ \frac{1}{2\ln 2} = \frac{1}{\ln(2^2)} = \frac{1}{\ln 4} \] Using the change of base formula again, $\frac{1}{\ln a} = \log_a e$. \[ \frac{1}{\ln 4} = \log_4 e \] 
Step 4: Final Answer 
The value of the derivative at $x=3$ is $\log_4 e$.