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
$\left({\frac {f}{g}}\right)'(x)={\frac {f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^{2}}}.$
Additionally, we will need our power rule for differentiation:
$\left(x^{n}\right)'\,=\,nx^{n-1},$ for $n\neq 0$,
as well as the derivative of natural log:
$\left(\ln x\right)'\,=\,{\frac {1}{x}}.$

 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  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 y\,=\,g\circ f(x)\,=\,g\left(f(x)\right).}

Step 2:
We can now apply all three advanced techniques. 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 f'(x)} , we must use both the quotient and product rule to find
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 \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
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 \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
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 y'=\displaystyle {\frac{x}{(x+5)(x-1)}\cdot\frac{(2x+5)x-(x^{2}+4x-5)(1)}{x^{2}}.}}

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

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