Find y' for y = ln ((x-1)³(x-2))/(x-3)
Find $y'$ for $y = \ln\dfrac{(x-1)^{3}(x-2)}{x-3}$.
Explanation of Method
Expand the logarithm first and the work becomes trivial.
Step 1: Expand using the logarithm laws
Products become sums, quotients become differences, and exponents become coefficients: $y = 3\ln(x-1) + \ln(x-2) - \ln(x-3)$.
Differentiating the original quotient directly would require the product and quotient rules inside a chain rule; after expansion, each term is a one-line derivative.
Step 2: Differentiate term by term
Each inner function has derivative $1$, so every term contributes simply the reciprocal of its argument: $y' = \frac{3}{x-1} + \frac{1}{x-2} - \frac{1}{x-3}$.
Step 3: Verification
\[x = 4:\quad y' = \frac{3}{3} + \frac{1}{2} - \frac{1}{1} = 1 + 0.5 - 1 = 0.5\]
\[\frac{y(4+10^{-5})-y(4-10^{-5})}{2\times 10^{-5}} \approx 0.5 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{3}{x-1} + \dfrac{1}{x-2} - \dfrac{1}{x-3}\,}\]
Remark
Remark. This is the technique of logarithmic differentiation in its purest form. The function itself is defined where the whole quotient is positive, but the derivative formula holds on every interval avoiding $x = 1, 2, 3$ — it is what one gets from $\ln|f|$, whose derivative $\dfrac{f'}{f}$ does not care about the sign of $f$.