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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} are differentiable functions, then
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (f\circ g)'(x) = f'(g(x))\cdot g'(x).}
The Product Rule: If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} are differentiable functions, then
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (fg)'(x) = f'(x)\cdot g(x)+f(x)\cdot g'(x).}
The Quotient Rule: If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} are differentiable functions and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x) \neq 0}  , then
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\frac{f}{g}\right)'(x) = \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^2}. }
Additionally, we will need our power rule for differentiation:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(x^n\right)'\,=\,nx^{n-1},} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n\neq 0} ,
as well as the derivative of natural log:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\ln x\right)'\,=\,\frac{1}{x}.}

 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
${\begin{array}{rcl}f'(x)&=&\displaystyle {\frac {\left((x+5)(x-1)\right)'x-(x+5)(x-1)(x)'}{x^{2}}}\\\\&=&\displaystyle {\frac {\left[x^{2}+4x-5\right]'x-(x^{2}+4x-5)(x)'}{x^{2}}}\\\\&=&\displaystyle {\frac {(2x+4)x-(x^{2}+4x-5)(1)}{x^{2}}}\\\\&=&\displaystyle {\frac {2x^{2}+4x-x^{2}-4x+5}{x^{2}}}\\\\&=&\displaystyle {\frac {x^{2}+5}{x^{2}}}.\end{array}}$

Step 3:
We can now use the chain rule to find
${\begin{array}{rcl}y'&=&\left(g\circ f\right)'(x)\\\\&=&g'\left(f(x)\right)\cdot f'(x)\\\\&=&\displaystyle {{\frac {x}{(x+5)(x-1)}}\cdot {\frac {x^{2}+5}{x^{2}}}}\\\\&=&\displaystyle {{\frac {x^{2}+5}{x^{3}+4x^{2}-5x}}.}\end{array}}$ 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
$y'=\displaystyle {{\frac {x}{(x+5)(x-1)}}\cdot {\frac {(2x+4)x-(x^{2}+4x-5)(1)}{x^{2}}}.}$

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

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