# 022 Exam 2 Sample A, Problem 1

Find the derivative of  ${\displaystyle y\,=\,\ln {\frac {(x+5)(x-1)}{x}}.}$

Foundations:
This problem is best approached through properties of logarithms. Remember that

${\displaystyle \ln(xy)=\ln x+\ln y,}$
and
${\displaystyle \ln \left({\frac {x}{y}}\right)=\ln x-\ln y,}$
You will also need to apply
The Chain Rule: If ${\displaystyle f}$ and ${\displaystyle g}$ are differentiable functions, then
${\displaystyle (f\circ g)'(x)=f'(g(x))\cdot g'(x).}$
Finally, recall that the derivative of natural log is
${\displaystyle \left(\ln x\right)'\,=\,{\frac {1}{x}}.}$

Solution:

Step 1:
We can use the log rules to rewrite our function as
${\displaystyle y\,=\,\ln(x+5)+\ln(x-1)-\ln(x).}$
Step 2:
We can differentiate term-by-term, applying the chain rule to the first two terms to find

${\displaystyle {\begin{array}{rcl}y'&=&\displaystyle {{\frac {1}{x+5}}\cdot (x+5)'+{\frac {1}{x-1}}\cdot (x-1)'+{\frac {1}{x}}}\\\\&=&\displaystyle {{\frac {1}{x+5}}+{\frac {1}{x-1}}+{\frac {1}{x}}}.\end{array}}}$
${\displaystyle y'\,=\,\displaystyle {{\frac {1}{x+5}}+{\frac {1}{x-1}}+{\frac {1}{x}}}.}$