022 Exam 2 Sample B, Problem 1

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

Foundations:
This problem is best approached through properties of logarithms. Remember that
$\ln(xy)=\ln x+\ln y,$
while
$\ln \left({\frac {x}{y}}\right)=\ln x-\ln y,$
and
$\ln \left(x^{n}\right)=n\ln x,$
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
${\begin{array}{rcl}y'&=&\displaystyle {\ln {\frac {(x+1)^{4}}{(2x-5)(x+4)}}}\\\\&=&4\ln(x+1)-\ln(2x-5)-\ln(x+4).\end{array}}$

Step 2:
We can differentiate term-by-term, applying the chain rule to each term to find
${\begin{array}{rcl}y'&=&\displaystyle {4\cdot {\frac {1}{x+1}}\cdot (x+1)'-{\frac {1}{2x-5}}\cdot (2x-5)'-{\frac {1}{x+4}}\cdot (x+4)'}\\\\&=&\displaystyle {{\frac {4}{x+1}}-{\frac {2}{2x-5}}-{\frac {1}{x+4}}}.\end{array}}$

Final Answer:
$y'\,=\,\displaystyle {{\frac {4}{x+1}}-{\frac {2}{2x-5}}-{\frac {1}{x+4}}}.$

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