Answer
$$ y'=3x^2-12x-79.$$
Work Step by Step
Taking the $\ln $ of both sides of the equation, we get $$\ln y= \ln ((x-1)(x-12)(x+7))$$ Then using the properties of $\ln $, we have $$\ln y= \ln (x-1)+\ln(x-12)+\ln(x+7).$$ Now taking the derivative for the above equation, we have $$\frac{y'}{y}= \frac{1}{ x-1}+ \frac{1}{ x-12}+\frac{1}{ x+7},$$ Hence $ y'$ is given by $$ y'=y\left( \frac{1}{ x-1}+ \frac{1}{ x-12}+\frac{1}{ x+7}\right)\\ = ((x-1)(x-12)(x+7))\left( \frac{1}{ x-1}+ \frac{1}{ x-12}+\frac{1}{ x+7}\right)\\=(x-12)(x+7)+(x-1)(x+7)+(x-1)(x-12)\\
=3x^2-12x-79.$$
