022 Exam 2 Sample A, Problem 1

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

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

 Solution:

Step 1:
We can use the log rules to rewrite our function as
$y\,=\,\ln(x+5)+\ln(x-1)-\ln(x).$

Step 2:
We can differentiate term-by-term, applying the chain rule to the first two terms to find
${\begin{array}{rcl}y'&=&\displaystyle {{\frac {1}{x+5}}\cdot (x+5)'+{\frac {1}{x-1}}\cdot (x-1)'+{\frac {1}{x}}}\\\\&=&\displaystyle {{\frac {1}{x+5}}+{\frac {1}{x-1}}+{\frac {1}{x}}}.\end{array}}$

Final Answer:
$y'\,=\,\displaystyle {{\frac {1}{x+5}}+{\frac {1}{x-1}}+{\frac {1}{x}}}.$

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