# 022 Exam 2 Sample B, Problem 1

Find the derivative of  ${\displaystyle y\,=\,\ln {\frac {(x+1)^{4}}{(2x-5)(x+4)}}.}$

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

${\displaystyle \ln(xy)=\ln x+\ln y,}$
while
${\displaystyle \ln \left({\frac {x}{y}}\right)=\ln x-\ln y,}$
and
${\displaystyle \ln \left(x^{n}\right)=n\ln x,}$
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

${\displaystyle {\begin{array}{rcl}y&=&\displaystyle {\ln {\frac {(x+1)^{4}}{(2x-5)(x+4)}}}\\\\&=&4\ln(x+1)-\ln(2x-5)-\ln(x+4).\end{array}}}$

Step 2:
We can differentiate term-by-term, applying the chain rule to each term to find

${\displaystyle {\begin{array}{rcl}y'&=&\displaystyle {4\cdot {\frac {1}{x+1}}\cdot (x+1)'-{\frac {1}{2x-5}}\cdot (2x-5)'-{\frac {1}{x+4}}\cdot (x+4)'}\\\\&=&\displaystyle {{\frac {4}{x+1}}-{\frac {2}{2x-5}}-{\frac {1}{x+4}}}.\end{array}}}$
${\displaystyle y'\,=\,\displaystyle {{\frac {4}{x+1}}-{\frac {2}{2x-5}}-{\frac {1}{x+4}}}.}$