Using logarithmic differentiation or otherwise, differentiate $y = (x −1) (x − 2) (x − 3$)? And show that $y' = 3x^2-12x+11$ So far this is my method:
\begin{align*}
y = (x-1)(x-2)(x-3) & \Longleftrightarrow \ln(y) = \ln(x-1) + \ln(x-2) + \ln(x-3)\\\\
& \Longleftrightarrow \frac{y'}{y} = \frac{1}{x-1} + \frac{1}{x-2} + \frac{1}{x-3}\\\\
& \Longleftrightarrow 
y' = y\left(\frac{1}{x-1} + \frac{1}{x-2} + \frac{1}{x-3}\right)
\end{align*}

The product rule is straightforward.
$$ y= (x-1)(x-2)(x-3)$$
$$ y' = (x-2)(x-3)+(x-1)(x-3)+(x-1)(x-2)=$$
$$ x^2-5x+6 + x^2-4x+3+x^2-3x+2 =$$
$$ 3x^2-12x+11$$
