Differentiate the following function $f(x)$ using the quotient rule.
", "name": "Luis's copy of Quotient rule - differentiate linear over quadratic", "metadata": {"notes": "\n\t\t1/08/2012:
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\n\t\tAdded description.
\n\t\tChecked calculation. OK.
\n\t\tAdded information about Show steps. Altered to 0 marks lost rather than 1.
\n\t\tChanged std rule set to include !noLeadingMinus, so polynomials don't change order. Got rid of a redundant ruleset.
\n\t\tImproved display in various places.
\n\t\tAdded condition that numbers have to be inout as fractions or integers - added decimal point to forbidden strings.
\n\t\t\n\t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "
The derivative of $\\displaystyle \\frac{ax+b}{cx^2+dx+f}$ is $\\displaystyle \\frac{g(x)}{(cx^2+dx+f)^2}$. Find $g(x)$.
"}, "ungrouped_variables": ["a", "c", "b", "d", "f", "s2", "s1", "det", "c1"], "parts": [{"prompt": "\n\t\t\t\\[\\simplify[std]{f(x) = ({a} * x+{b})/({c}x^2+{d}x+{f})}\\]
You are given that \\[\\frac{df}{dx}=\\simplify[std]{g(x)/({c}x^2+{d}x+{f})^2}\\]
for a polynomial $g(x)$. You are asked to find $g(x)$
$g(x)=\\;$[[0]]
\n\t\t\tInput numbers as fractions or integers and not as decimals.
\n\t\t\tClick on Show steps for more information. You will not lose any marks by doing so.
\n\t\t\t", "type": "gapfill", "marks": 0, "showCorrectAnswer": true, "gaps": [{"checkvariablenames": false, "showCorrectAnswer": true, "answer": "{-c*a}x^2+{-2*b*c}x+{a*f-b*d}", "vsetrange": [0, 1], "answersimplification": "std", "expectedvariablenames": [], "vsetrangepoints": 5, "type": "jme", "scripts": {}, "checkingtype": "absdiff", "notallowed": {"partialCredit": 0, "message": "Input numbers as fractions or integers and not as decimals.
", "showStrings": false, "strings": ["."]}, "marks": 3, "checkingaccuracy": 0.001, "showpreview": true}], "stepsPenalty": 0, "scripts": {}, "steps": [{"prompt": "The quotient rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u/v,x,1) = (v * Diff(u,x,1) - u * Diff(v,x,1))/v^2}\\]
The quotient rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u/v,x,1) = (v * Diff(u,x,1) - u * Diff(v,x,1))/v^2}\\]
For this example:
\n\t \n\t \n\t \n\t\\[\\simplify[std]{u = ({a}x+{b})}\\Rightarrow \\simplify[std]{Diff(u,x,1) = {a}}\\]
\n\t \n\t \n\t \n\t\\[\\simplify[std]{v = ({c} * x^2+{d}x+{f})} \\Rightarrow \\simplify[std]{Diff(v,x,1) = {2*c}x+{d}}\\]
\n\t \n\t \n\t \n\tHence on substituting into the quotient rule above we get:
\n\t \n\t \n\t \n\t\\[\\begin{eqnarray*} \\frac{df}{dx}&=&\\simplify[std]{({a}({c}x^2+{d}x+{f})-({2*c}x+{d})({a}x+{b}))/({c}x^2+{d}x+{f})^2}\\\\\n\t \n\t &=&\\simplify[std]{({a*c}x^2+{a*d}x+{a*f}-{2*c*a}x^2-{a*d+2*c*b}x+{d*b})/({c}x^2+{d}x+{f})^2}\\\\\n\t \n\t &=&\\simplify[std]{({-c*a}x^2+{-2*b*c}x+{a*f-d*b})/({c}x^2+{d}x+{f})^2}\n\t \n\t \\end{eqnarray*}\\]
Hence $g(x)=\\simplify[std]{{-c*a}x^2+{-2*b*c}x+{a*f-d*b}}$