022 Exam 2 Sample A, Problem 1

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

Foundations:
This problem requires several advanced rules of differentiation. In particular, you need
The Chain Rule: If $f$ and $g$ are differentiable functions, then
$(f\circ g)'(x)=f'(g(x))\cdot g'(x).$
The Product Rule: If $f$ and $g$ are differentiable functions, then
$(fg)'(x)=f'(x)\cdot g(x)+f(x)\cdot g'(x).$
The Quotient Rule: If $f$ and $g$ are differentiable functions and $g(x)\neq 0$ , then
$\left({\frac {f}{g}}\right)'(x)={\frac {f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^{2}}}.$

 Solution:

Step 1:
We need to identify the composed functions in order to apply the chain rule. Note that if we set $g(x)\,=\,\ln x$, and
$f(x)\,=\,{\frac {(x+5)(x-1)}{x}},$
we then have  $y\,=\,g\circ f(x)\,=\,g\left(f(x)\right).$

Step 2:
We can now apply all three advanced techniques. For $f'(x)$, we must use both the quotient and product rule to find
${\begin{array}{rcl}f'(x)&=&\displaystyle {\frac {\left((x+5)(x-1)\right)'x-(x+5)(x-1)(x)'}{x^{2}}}\\\\&=&\displaystyle {\frac {\left[x^{2}+4x-5\right]'x-(x^{2}+4x-5)(x)'}{x^{2}}}\\\\&=&\displaystyle {\frac {(2x+5)x-(x^{2}+4x-5)(1)}{x^{2}}}\\\\&=&\displaystyle {\frac {2x^{2}-5x-x^{2}-4x+5}{x^{2}}}\\\\&=&\displaystyle {\frac {x^{2}-9x+5}{x^{2}}}.\end{array}}$

Step 3:
We can now use the chain rule to find
${\begin{array}{rcl}y'&=&\left(g\circ f\right)'(x)\\\\&=&g'\left(f(x)\right)\cdot f'(x)\\\\&=&\displaystyle {{\frac {x}{(x+5)(x-1)}}\cdot {\frac {x^{2}-9x+5}{x^{2}}}}\\\\&=&\displaystyle {{\frac {x^{2}-9x+5}{x^{3}+4x^{2}-5x}}.}\end{array}}$ Note that many teachers do not prefer a cleaned up answer, and may request that you do not simplify. In this case, we could write the answer as
$y'=\displaystyle {{\frac {x}{(x+5)(x-1)}}\cdot {\frac {(2x+5)x-(x^{2}+4x-5)(1)}{x^{2}}}.}$

Final Answer:
$y'\,=\,\displaystyle {{\frac {x^{2}-9x+5}{x^{3}+4x^{2}-5x}}.}$

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